Quantitative chemistry Flashcards
(made w/ freesciencelessons) - left off at "Using concentration of solution 2 (Triple)" (109 cards)
1
Q
What is the law of conservation of mass?
A
The idea that no atoms are lost or made during a chemical reaction, therefore, the mass of the products equals the mass of the reactants and the number of atoms in each element is the same.
2
Q
What is a reactant?
A
The starting substances in a reaction.
3
Q
What is a product?
A
The chemicals formed in a reactant.
4
Q
Question:
92g of sodium reacts with 32g of oxygen. Calculate the mass of sodium oxide produced.
4Na + O₂ ⭢ 2Na₂O
A
92 + 32 = 124g
Total mass of product(s) = total mass of reactant(s)
5
Q
Question:
A mass of calcium carbonate reacted to produce 112g of calcium oxide and 88g of carbon dioxide. Calculate the mass of calcium carbonate that reacted.
CaCO₃ ⭢ CaO + CO₂
A
112 + 88 = 200g
Total mass of reactant(s) = total mass of product(s)
6
Q
Question:
A mass of magnesium oxide reacts with 73g of hydrogen chloride to produce 95g of magnesium chloride and 18g of water. Calculate the mass of magnesium oxide in the reaction.
MgO + 2HCl ⭢ MgCl₂ + H₂O
A
95 + 18 = 113g
113 - 73 = 40g
Total mass of product(s) = total mass of reactant(s)
7
Q
How [when are] are ionic compounds formed?
A
When metals react with non-metals.
8
Q
What are ions?
A
Atoms of an element with an overall charge that are formed when atoms gain or lose electrons.
9
Q
What ions do metals / [most] non-metals form?
A
- Metals form positive ions.
- Most non-metals form negative ions.
10
Q
What often correlates with the charge of an ion? What does not follow this pattern? Why?
A
The group number of the metals;
* transition metals as they can have [form] several different [ions] charges
* non-metal ionic compounds as they consist of several nonmetal atoms.
11
Q
What are the charges of the non-metal ionic compounds?
A
Hydroxide: OH⁻ (1-)
Nitrate: NO₃⁻ (1-)
Sulfate: SO₄²⁻ (2-)
Carbonate: CO₃²⁻ (2-)
Ammonium: NH₄⁺ (1+)
12
Q
Question:
Find the formula, using the charges, for sodium chloride.
A
Na⁺ + Cl⁻ ⭢ NaCl
13
Q
Question:
Find the formula, using the charges, for sodium oxide.
A
Na⁺ + O²⁻ ⤚ (Na × 2) ⭢ Na₂O
14
Q
Question:
Find the formula, using the charges, for magnesium iodide.
A
Mg²⁺ + I⁻ ⤚ (I × 2) ⭢ MgI₂
15
Q
Question:
Find the formula, using the charges, for lithium carbonate.
A
Li⁺ + CO₃²⁻ ⤚ (Li × 2) ⭢ Li₂CO₃
16
Q
Question:
Find the formula, using the charges, for calcium hydroxide.
A
Ca²⁺ + OH⁻ ⤚ (OH × 2) ⭢ Ca(OH)₂
17
Q
Question:
Find the formula, using the charges, for magnesium nitrate.
A
Mg²⁺ + NO₃⁻ ⤚ (NO₃ × 2) ⭢ Mg(NO₃)₂
18
Q
Question:
Balance the equation:
Calcium oxide + hydrochloric acid ⭢ Calcium chloride + water
CaO + HCl ⭢ CaCl₂ + H₂O
A
CaO + 2HCl ⭢ CaCl₂ + H₂O
19
Q
Question:
Balance the equation:
Iron oxide + carbon monoxide ⭢ Iron + carbon dioxide
Fe₂O₃ + CO ⭢ Fe + CO₂
A
Fe₂O₃ + 3CO ⭢ 2Fe + 3CO₂
20
Q
What is the Relative Atomic Mass (Ar)?
A
The average mass of the atoms of an element, taking into account naturally occurring isotopes, compared with carbon-12 (which is given a mass of exactly 12).
21
Q
What is the Relative Formula Mass (Mr) of a compound?
A
The sum of the relative atomic masses (Ar) of the atoms in the numbers shown in the formula.
22
Q
What are two things to watch out for when calculating the Relative Formula Mass (Mr)?
A
- They have no units
- Do not take into account coefficients (big numbers) when calculating the Mr.
23
Q
Question:
Calculate the Mr of CaSO₄
(Ar of Ca = 40), (Ar of S = 32), (Ar of O = 16)
A
Mr = 40 + 32 + (16 × 4) = 136
24
Q
Question:
Calculate the Mr of Mg(OH)₂
(Ar of Mg = 24), (Ar of O = 16), (Ar of H = 1)
A
Mr = 24 + (16 × 2) + (1 × 2) = 58
25
What is the equation for calculating the percentage composition (percentage by mass)?
% composition: 100 (total Ar of element ÷ total Mr of compound)
26
# Question:
Calculate the percentage by mass of calcium (Ca) in calcium chloride (CaCl₂).
Ar of Ca = 40
Mr of CaCl₂ = 111
Percentage by mass of Ca = (40 ÷ 111) × 100 = (0.3603) × 100 = _36%_
| % composition: 100 (total Ar of element ÷ total Mr of compound)
27
# Question:
Calculate the percentage by mass of hydrogen (H) in calcium chloride (CH₄).
Ar of H = 1
Mr of CH₄ = 16
Percentage by mass of H = ((4 × 1) ÷ 16) × 100 = (0.25) × 100 = _25%_
| % composition: 100 (total Ar of element ÷ total Mr of compound)
28
# Question:
Calculate the percentage by mass of sodium (Na) in sodium sulphate (Na₂SO₄).
Mr of Na₂SO₄ = (23 × 2) + 32 + (16 × 4) = 142
Ar of Na = 23
Percentage by mass of Na = ((23 × 2) ÷ 142) × 100 = (0.3239) × 100 = _32.4%_
| % composition: 100 (total Ar of element ÷ total Mr of compound)
29
What is the equation for calculating the number of moles of an element?
| [Higher tier]
Number of moles (mol) = Mass (g) ÷ Relative atomic mass *or* Ar
30
# Question:
You are given a sample of magnesium with a mass of 48g. How many moles of magnesium have you been given?
| [Higher tier]
Number of moles = 48g ÷ 24 = _2 mol_
| # of moles (mol) = Mass (g) ÷ Relative atomic mass *or* Ar
31
# Question:
You are given 120g of calcium. How many moles of calcium have you been given?
| [Higher tier]
Number of moles = 120g ÷ 40 = _3 mol_
| # of moles (mol) = Mass (g) ÷ Relative atomic mass *or* Ar
32
# Question:
A sample of rock contains 252g of iron. Calculate the number of moles of iron in the sample.
| [Higher tier]
Number of moles = 252g ÷ 56 = _4.5 mol_
| # of moles (mol) = Mass (g) ÷ Relative atomic mass *or* Ar
33
# Question:
You are given a sample of sulphur with a mass of 4064g. Calculate the number of moles of sulphur in the sample.
| [Higher tier]
Number of moles = 4064g ÷ 32 = _127 mol_
| # of moles (mol) = Mass (g) ÷ Relative atomic mass *or* Ar
34
What is the equation for calculating the number of moles of a compound?
| [Higher tier]
Number of moles (mol) = Mass (g) ÷ Relative formula mass *or* Mr
35
# Question:
You are given a sample of calcium carbonate (CaCO₃) with a mass of 300g. Calculate the number of moles of calcium carbonate in the sample.
| [Higher tier]
Mr of CaCO₃ = 40 + 12 + (16 × 3) = 100
Number of moles = 300g ÷ 100 = _3 mol_
| # of moles (mol) = Mass (g) ÷ Relative formula mass *or* Mr
36
# Question:
You are given 380g of magnesium chloride (MgCl₂). How many moles of magnesium chloride have you been given?
| [Higher tier]
Mr of MgCl₂ = 24 + (35.5 × 2) = 95
Number of moles = 380g ÷ 95 = _4 mol_
| # of moles (mol) = Mass (g) ÷ Relative formula mass *or* Mr
37
# Question:
You are given a sample of lithium sulfate (Li₂SO₄) with a mass of 990g. Calculate the number of moles of lithium sulfate in the sample.
| [Higher tier]
Mr of Li₂SO₄ = (7 × 2) + 32 + (16 × 4) = 110
Number of moles = 990g ÷ 110 = _9 mol_
| # of moles (mol) = Mass (g) ÷ Relative formula mass *or* Mr
38
# Question:
You are given 64.5g of beryllium hydroxide (Be(OH)₂). Calculate the number of moles of beryllium hydroxide that you have been given.
| [Higher tier]
Mr of Be(OH)₂ = 9 + (16 × 2) + (1 × 2) = 43
Number of moles = 64.5g ÷ 43 = _1.5 mol_
| # of moles (mol) = Mass (g) ÷ Relative formula mass *or* Mr
39
Rearrange the equation for calculating the number of moles of a compound and an element to make mass the subject.
| [Higher tier]
* Mass (g) = number of moles (mol) × Relative formula mass *or* Mr
* Mass (g) = number of moles (mol) × Relative atomic mass *or* Ar
40
# Question:
Calcuate the mass of four moles of sodium chloride (NaCl).
| [Higher tier]
Mr of NaCl = 23 + 35.5 = 58.5
Mass = 4 mol × 58.5 = _234g_
| Mass (g) = number of moles (mol) × Relative formula mass *or* Mr
41
# Question:
Calculate the mass of three moles of potassium oxide (K₂O).
| [Higher tier]
Mr of K₂O = (39 × 2) + 16 = 94
Mass = 3 mol × 94 = _282g_
| Mass (g) = number of moles (mol) × Relative formula mass *or* Mr
42
# Question:
Calculate the mass of 0.1 moles of caesium nitrate (CsNO₃).
| [Higher tier]
Mr of CsNO₃ = 133 + 14 + (16 × 3) = 195
Mass = 0.1 mol × 195 = _19.5g_
| Mass (g) = number of moles (mol) × Relative formula mass *or* Mr
43
# Question:
Calculate the mass of 5 moles of coper sulfate (CuSO₄).
| [Higher tier]
Mr of CuSO₄ = 63.5 + 32 + (16 × 4) = 159.5
Mass = 5 mol × 159.5 = _797.5g_
| Mass (g) = number of moles (mol) × Relative formula mass *or* Mr
44
#Question:
2g of hydrogen reacts with 71g of chlorine to make 73g of hydrogen chloride. Using this information, balance the equation.
H₂ + Cl₂ ⭢ HCl
| [Higher tier]
Moles of hydrogen = 2g ÷ 2 = 1 mol
Moles of chlorine = 71g ÷ 71 = 1 mol
Moles of hydrogen chloride = 73g ÷ 36.5 = 2 mol
| 1 ÷ 1 = 1 | 1 ÷ 1 = 1 | 2 ÷ 1 = 2 |
H₂ + Cl₂ ⭢ **_2_**HCl
##Footnote
Number of moles (mol) = Mass (g) ÷ Ar *or* Mr
Smallest coefficients = coefficients ÷ smallest coefficient
45
# Question:
54g of aluminium reacts with 216g of iron (II) oxide, forming 102g of aluminium oxide and 168g of iron. Using this information, balance the equation.
Al + FeO ⭢ Al₂O₃ + Fe
| [Higher tier]
Moles of aluminium = 54g ÷ 27 = 2 mol
Moles of iron (II) oxide = 216g ÷ (56 + 16) = 3 mol
Moles of aluminium oxide = 102g ÷ ((27 × 2) + (16 × 3)) = 1 mol
Moles of iron = 168g ÷ 56 = 3 mol
| 2 ÷ 1 = 2 | 3 ÷ 1 = 3 | 1 ÷ 1 = 1 | 3 ÷ 1 = 3 |
**_2_**Al + **_3_**FeO ⭢ Al₂O₃ + **_3_**Fe
##Footnote
Number of moles (mol) = Mass (g) ÷ Ar *or* Mr
Smallest coefficients = coefficients ÷ smallest coefficient
46
# Question:
1248g of barium chloride reacts with 684g of aluminium sulfate, forming 1398g of barium sulfate and 534g of aluminium chloride. Using this information, balance the equation.
BaCl₂ + Al₂(SO₄)₃ ⭢ BaSO₄ + AlCl₃
| [Higher tier]
Moles of barium chloride = 1248g ÷ (137 + (35.5 × 2)) = 6 mol
Moles of aluminium sulfate = 684g ÷ ((27 × 2) + (32 × 3) + 3(16 × 4)) = 2 mol
Moles of barium sulfate = 1398g ÷ (137 + 32 + (16 × 4)) = 6 mol
Moles of aluminium chloride = 534g ÷ (27 + (35.5 × 3)) = 4 mol
| 6 ÷ 2 = 3 | 2 ÷ 2 = 1 | 6 ÷ 2 = 3 | 4 ÷ 2 = 2 |
**_3_**BaCl₂ + Al₂(SO₄)₃ ⭢ **_3_**BaSO₄ + **_2_**AlCl₃
##Footnote
Number of moles (mol) = Mass (g) ÷ Ar *or* Mr
Smallest coefficients = coefficients ÷ smallest coefficient
47
# Question:
24g of magnesium reacts with 16g of oxygen, forming 40g of magnesium oxide. Using this information, balance the equation.
Mg + O₂ ⭢ MgO
| [Higher tier]
Moles of magnesium = 24g ÷ 24 = 1 mol
Moles of oxygen = 16g ÷ (16 × 2) = 0.5 mol
Moles of magnesium oxide = 40g ÷ (24 + 16) = 1 mol
| 1 ÷ 0.5 = 2 | 0.5 ÷ 0.5 = 1 | 1 ÷ 0.5 = 2 |
**_2_**Mg + O₂ ⭢ **_2_**MgO
##Footnote
Number of moles (mol) = Mass (g) ÷ Ar *or* Mr
Smallest coefficients = coefficients ÷ smallest coefficient
48
What is Avogadro's constant?
| [Higher tier]
The number of atoms, molecules or ions in a mole of a given substance.
49
What is the value of Avogadro's constant?
| [Higher tier]
6.02 × 10²³
50
# Question:
Calculate the number of moles of atoms in one mole of water molecules (H₂O).
1 molecule of H₂O = 3 atoms
∴ 1 mole of H₂O = _3 moles of atoms_
51
# Question:
Calculate the number of moles of atoms in one mole of methane (CH₄).
| [Higher tier]
1 molecule of CH₄ = 5 atoms
∴ 1 mole of CH₄ = _5 moles of atoms_
52
# Question:
Calculate the number of moles of atoms in one mole of calcium hydroxide (Ca(OH)₂).
| [Higher tier]
1 molecule of Ca(OH)₂ = 5 atoms
∴ 1 mole of Ca(OH)₂ = _5 moles of atoms_
53
# Question:
Calculate the number of atoms in one mole of hydrogen chloride (HCl).
| [Higher tier]
1 molecule of HCl = 2 atoms
1 mole of HCl = 6.02 × 10²³
Number of atoms in one mole 2 × 6.02 × 10²³ = _1.204 × 10²⁴_
54
# Question:
Calculate the number of atoms in one mole of sodium oxide (Na₂O).
| [Higher tier]
Number of atoms in 1 molecule of Na₂O = 3
1 mole = 6.02 × 10²³
Number of atoms in one mole = 3 × 6.02 × 10²³ = _1.806 × 10²⁴_
55
# Question:
Calculate the number of atoms in 48g of magnesium (Mg).
| [Higher tier]
Moles (mol) = 48g ÷ 24 = 2 mol
1 atom in one element of magnesium.
Number of atoms = 2 × 6.02 × 10²³
| # of moles (mol) = Mass (g) ÷ Relative atomic mass *or* Ar
56
# Question:
Calculate the number of atoms in 28g of lithium (Li).
| [Higher tier]
Moles (mol) = 28g ÷ 7 = 4 mol
1 atom in one element of lithium.
Number of atoms = 4 mol × 6.02 × 10²³ = _2.408 × 10²⁴_
| # of moles (mol) = Mass (g) ÷ Relative atomic mass *or* Ar
57
# Question:
Calculate the number of atoms in 56g of calcium oxide (CaO).
| [Higher tier]
Mole (mol) = 56g ÷ (40 + 16) = 1 mol
2 atoms in one molecule of calcium oxide.
Number of atoms = 2 × 6.02 × 10²³ = _1.204 × 10²⁴_
| # of moles (mol) = Mass (g) ÷ Relative formula mass *or* Mr
58
# Question:
Calculate the number of atoms in 54g of water (H₂O).
| [Higher tier]
Moles = 54g ÷ ((1 × 2) + 16) = 3 mol
3 atoms in one molecule of water.
3 × 3 = 9
Number of atoms = 9 × 6.02 × 10²³ = _5.418 × 10²⁴_
| # of moles (mol) = Mass (g) ÷ Relative formula mass *or* Mr
59
What are the steps to calculating the masses of products or reactants using reacting masses in a chemical equation?
| [Higher tier]
1. Identify the known and unknown substances from the question.
2. Calculate Mr or Ar of the known substance using your periodic table.
3. Calculate moles of known substance using Mr (or Ar) and the given mass.
4. Workout the moles of the unknown substance using the ratio, which are the coefficients.
5. Calculate the Mr or Ar of the unknown substance using your periodic table.
6. Calculate the mass of the unknown substance using the moles and Mr (or Ar).
60
# Question:
Calculate the mass of magnesium chloride that could be produced from 72g of magnesium. Assume that the chlorine is unlimited.
Mg + Cl₂ ⭢ MgCl₂
| [Higher tier]
1. _Identify known & unknown substance._
Known: Mg | Unknown: MgCl₂
2. _Ar or Mr of known substance._
Ar of Mg = 24
3. _Moles of known substance._
Moles (mol) = 72g ÷ 24 = 3 mol
4. _Moles of unknown substance from ratio._
Ratio = 1:1 | Moles (mol) = 3 moles
5. _Mr or Ar of unknown substance._
Mr of MgCl₂ = 95
6. _Mass of unknown substance._
Mass (g) = 3 mol × 95 = _285g_
## Footnote
Moles (mol) = mass (g) ÷ relative atomic mass *or* Ar
Moles (mol) = mass (g) ÷ relative formula mass *or* Mr
61
# Question:
Calculate the mass of calcium sulfate that could be produced from 80g of calcium. Assume that the sulfuric acid is unlimited.
Ca + H₂SO₄ ⭢ CaSO₄ + H₂
| [Higher tier]
1. _Identify known & unknown substance._
Known: Ca | Unknown: CaSO₄
2. _Ar or Mr of known substance._
Ar of Ca = 40
3. _Moles of known substance._
Moles (mol) = 80g ÷ 40 = 2 mol
4. _Moles of unknown substance from ratio._
Ratio = 1:1 | Moles (mol) = 2 moles
5. _Mr or Ar of unknown substance._
Mr of CaSO₄ = 136
6. _Mass of unknown substance_
Mass (g) = 2 mol × 136 = _272g_
## Footnote
Moles (mol) = mass (g) ÷ relative atomic mass *or* Ar
Moles (mol) = mass (g) ÷ relative formula mass *or* Mr
62
# Question:
Calculate the mass of calcium carbonate that we would need to produce 224g of calcium oxide.
CaCO₃ ⭢ CaO + CO₂
| [Higher tier]
1. _Identify known & unknown substance._
Known: CaO | Unknown: CaCO₃
2. _Mr or Ar of known substance._
Mr of CaO = 56
3. _Moles of known substance_
Moles (mol) = 224g ÷ 56 = 4 mol
4. _Moles of unknown substance from ratio._
Ratio: 1:1 | Moles (mol) = 4 moles
5. _Mr or Ar of unknown substance._
Mr of CaCO₃ = 100
6. _Mass of unknown substance._
Mass (g) = 4 mol × 100 = _400g_
## Footnote
Moles (mol) = mass (g) ÷ relative formula mass *or* Mr
63
# Question:
Calculate the mass of magnesium chloride that could be produced from 146g of hydrochloric acid. Assume that the magnesium hydroxide is unlimited.
Mg(OH)₂ + 2HCl ⭢ MgCl₂ + H₂O
| [Higher tier]
1. _Identify known and unknown substance._
Known: HCl | Unknown: MgCl₂
2. _Mr or Ar of known substance._
Mr of HCl = 36.5
3. _Moles of known substance_
Moles (mol) = 146g ÷ 36.5 = 4 mol
4. _Moles of unknown substance from ratio._
Ratio = 2:1 | Moles (mol) = 4 ÷ (2 ÷ 1) = 2 moles
5. _Mr or Ar of unknown substance._
Mr of MgCl₂ = 95
6. _Mass of unknown substance._
Mass (g) = 2 mol × 95 = _190g_
## Footnote
Moles (mol) = mass (g) ÷ relative formula mass *or* Mr
64
# Question:
Calculate the mass of sodium sulfate that could be produced from 240g of sodium hydroxide. Assume that the sulfuric acid is unlimited.
2NaOH + H₂SO₄ ⭢ Na₂SO₄ + 2H₂O
| [Higher tier]
1. _Identify known and unknown substance._
Known: NaOH | Unknown: Na₂SO₄
2. _Mr or Ar of known substance._
Mr of NaOH = 40
3. _Moles of known substance_
Moles (mol) = 240g ÷ 40 = 6 mol
4. _Moles of unknown substance from ratio._
Ratio = 2:1 | Moles (mol) = 6 ÷ (2 ÷ 1) = 3 moles
5. _Mr or Ar of unknown substance._
Mr of Na₂SO₄ = 142
6. _Mass of unknown substance._
Mass (g) = 3 mol × 142 = _426g_
## Footnote
Moles (mol) = mass (g) ÷ relative formula mass *or* Mr
65
# Question:
Calculate the mass of hydrogen peroxide that could produce 64g of oxygen.
2H₂O₂ ⭢ 2H₂O + O₂
| [Higher tier]
1. _Identify known and unknown substance._
Known: O₂ | Unknown: H₂O₂
2. _Mr or Ar of known substance._
Ar of O₂ = 16 × 2 = 32
3. _Moles of known substance_
Moles (mol) = 64g ÷ 32 = 2 mol
4. _Moles of unknown substance from ratio._
Ratio = 1:2 | Moles (mol) = 2 × 2 = 4 moles
5. _Mr or Ar of unknown substance._
Mr of H₂O₂ = 34
6. _Mass of unknown substance._
Mass (g) = 4 mol × 34 = _136g_
## Footnote
Moles (mol) = mass (g) ÷ relative atomic mass *or* Ar
Moles (mol) = mass (g) ÷ relative formula mass *or* Mr
66
# Question:
Nitrogen and hydrogen form ammonia shown by the following equation:
N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
Calculate the mass of nitrogen needed to form 6.8 tonnes of ammonia.
| [Higher tier]
1. _Identify known and unknown substance._
Known: NH₃ | Unknown: N₂
2. _Mr of known substance._
Mr of NH₃ = 17
3. _Convert mass into grams_
1 tonne = 1000kg = 1 × 10⁶g | Mass = 6.8 tonnes ⤚ (×1,000,000) ⭢ 6,800,000g
4. _Moles of known substance._
Moles (mol) = (6.8 × 10⁶)g ÷ 17 = 400,000 mol
5. _Moles of unknown from ratio._
Ratio = 2:1 | Moles = 400,000 ÷ (2 ÷ 1) = 200,000 moles
6. _Mr of unknown substance_
Mr of N₂ = 28
7. _Mass of unknown substance._
Mass (g) = 200,000 mol × 28 = _5,600,000g_ (5.6 tonnes)
## Footnote
Moles (mol) = mass (g) ÷ relative formula mass *or* Mr
67
What is a limited reactant (*or* reagent)?
| [Higher tier]
The reactant (in a chemical equation) that is completely used up, limiting the amount of products produced.
68
What is the excess reactant (*or* reagent)?
| [Higher tier]
The reactant or reagent that is left after a chemical reaction is complete.
69
What are the steps to calculating the masses of products or reactants using limited reactants in a chemical equation?
| [Higher tier]
1. Identify and ignore, for the rest of your calculations, the excess reactant.
2. Identify the known and unknown substance from the limited reactant and what you're trying to find.
3. Find the moles of product from moles of limited reactant (they are the same).
4. Find the mass of the product using moles and Mr or Ar.
70
# Question:
How many moles of zinc iodide would be produced if we used 0.5 moles of zinc and 1 mole of iodine? Calculate the mass of product.
Zn + I₂ ⭢ ZnI₂
| [Higher tier]
1. _Identify and ignore the excess reactant._
Given moles: 1 > 0.5 | Excess reactant = I₂
2. _Identify known and unknown substance._
Known: Zn | Unkown: ZnI₂
3. _Find moles of product from limited reactant._
Moles of Zn = moles of ZnI₂ = 0.5 mol
4. _Mass of product._
Mass (g) = 0.5 mol × (65 + (127 × 2)) = _159.5g_
## Footnote
Mass (g) = number of moles (mol) × relative formula mass *or* Mr
71
# Question:
How many moles of sodium chloride will be produced if we used 1 mole of sodium hydroxide and 0.25 moles of hydrochloric acid? Calculate the mass of sodium chloride produced.
NaOH + HCl ⭢ NaCl + H₂O
| [Higher tier]
1. _Identify and ignore the excess reactant._
Given moles: 1 > 0.25 | Excess = NaOH
2. _Identify known and unknown substance._
Known: NaOH | Unknown: NaCl
3. _Moles of product from limited reactant._
Moles of NaOH = moles of NaCl | Moles = 0.25 mol
4. _Mass of product._
Mass (g) = 0.25 mol × (23 + 35.5) = _14.625g_
## Footnote
Mass (g) = number of moles (mol) × relative formula mass *or* Mr
72
# Question:
How many moles of copper will be produced if we used 0.5 moles of copper sulphate and 1 mole of magnesium? Calculate the mass of copper produced.
CuSO₄ + Mg ⭢ MgSO₄ + Cu
| [Higher tier]
1. _Identify and ignore the excess reactant._
Given moles: 0.5 > 1 | Excess = Mg
2. _Identify known and unknown substance._
Known: CuSO₄ | Unknown: Cu
3. _Moles of product from limited reactant._
Moles of CuSO₄ = moles of Cu | Moles = 0.5 mol
4. _Mass of product._
Mass (g) = 0.5 mol × 63.5 = _31.75g_
## Footnote
Mass (g) = number of moles (mol) × relative atomic mass *or* Ar
73
What is meant by concentration; what is the unit?
| [Foundation & Higher tier]
Concentration tells us the mass of a solute in a given volume of solution; g/dm³.
74
What is a solute?
A chemical that is dissolved in a solvent.
75
What is the equation for calculating the concentration of a solution?
| (not in specification but worth learning)
Concentration (g/dm³) = mass (g) ÷ volume (dm³)
76
# Question:
200g of a chemical is dissolved in water to a final volume of 1 dm³. Calculate the concentration of the solution.
Concentration = 200g ÷ 1 dm³ = _200g/dm³_
| Concentration (g/dm³) = mass (g) ÷ volume (dm³)
77
# Question:
150g of a chemical is dissolved in water to a final volume of 0.5 dm³. Calculate the concentration of the solution.
Concentration (g/dm³) = 150g ÷ 0.5 dm³ = _300 g/dm³_
| Concentration (g/dm³) = mass (g) ÷ volume (dm³)
78
# Quetsion:
Calculate the mass of a chemical needed to dissolve in a final volume of 0.4 dm³ to give a final concentration of 600 g/dm³.
Mass (g) = 600 g/dm³ × 0.4 dm³ = _240g_
| Mass (g) =| Concentration (g/dm³) × volume (dm³)
79
# Question:
Calculate the final volume of a solution containing 200g of a chemical with a concentration of 800 g/dm³.
Volume (dm³) = 200g ÷ 800g/dm³ = 0.25 dm³
| Volume (dm³) = mass (g) ÷ concentration (g/dm³)
80
What 2 factors affect the concentration?
| [Higher tier]
* Mass of the solute: increase in mass = increase in concentration (if volume of solution kept the same).
* Volume of the solution: increase in volume = decrease in concentration (if mass of solute kept the same).
81
What is yield?
| **Triple only**
The mass of product that a chemical reaction produces.
82
What is the problem with calculating the yield, in practice?
| **Triple only**
It is not always possible to achieve 100% yield in a chemical reaction.
83
What are the reasons why you cannot always achieve 100% yield in a chemical reaction? (3)
| **Triple only**
1. Some of the product may be lost when it is separated from the reaction mixture.
2. Some of the reactants may react in different ways to the expected reaction so we do not get the product we expect.
3. Reversible reactions may not go to completion.
84
What is the equation for calculating the percentage yield?
| **Triple only**
Percentage yield = 100 (mass of product actually made ÷ maximum theoretical mass of product)
85
What are the steps to calculating the percentage yield?
| **Triple only**
1. Find out the theoretical mass by calculating the mass of the product provided in the question, using reacting masses.
2. Calculate the percentage yield using your calculated maximum theoretical mass and the mass of the product given in the question.
86
# Question:
A scientist reacted 48g of magnesium and produced 150g of magnesium sulphate. Calculate the percentage yield.
Mg + H₂SO₄ ⭢ MgSO₄ + H₂
**Triple only**
Step 1:
1. _Identify known and unknown substances_
Known: Mg | Unknown: MgSO₄
2. _Mr of known and unknown substance._
Ar of Mg = 24 | Mr of MgSO₄ = 120
3. _Moles of known substance._
Moles (mol) = 48g ÷ 24 = 2 mol
4. _Mole of unknown from ratio._
Ratio = 1:1 | Moles (mol) = 2 mol
5. _Mass of unknown._
[Theoretical] mass (g) = 2 mol × 120 = _240 g_
Step 2:
Percentage yield = 150g ÷ 240g = 0.625 × 100 = _62.5%_
## Footnote
* Number of moles (mol) = mass (g) ÷ relative formula mass *or* Mr
* % yield = 100 (actual mass ÷ theoretical mass)
87
# Question:
The theoretical yield of beryllium chloride was 10.7g. If the reaction actually yields 4.5g, what was the percent yield?
Be + 2HCl ⭢ BeCl₂ + H₂
| **Triple only**
Percentage yield = (4.5g ÷ 10.7g) × 100 = 0.42 × 100 = _42%_
| % yield = 100 (actual mass ÷ theoretical mass)
88
# Question:
2a) A scientist began this reaction with 20 grams of lithium hydroxide and an unlimited amount of KCl.
What is the theoretical yield of lithium chloride (grams)?
LiOH + KCl ⭢ LiCl + KOH
2b) The reaction actually produced 6 grams of lithium chloride. What is the percent yield?
| **Triple only**
Part A:
1. _Identify known and unknown substances._
1Known: LiOH | Unknown: LiCl
2. _Mr of known and unknown substance._
Mr of LiOH = 24 | Mr of LiCl = 42.5
3. _Moles of known substance._
Moles (mol) = 20g ÷ 24 = 0.8 mol
4. _Mole of unknown from ratio._
Ratio = 1:1 | Moles (mol) = 0.8 mol
5. _Mass of unknown._
[Theoretical] mass (g) = 0.8 mol × 42.5 = _34 g_
Part B:
Percentage yield = 6g ÷ 34g = 0.176 × 100 = _17.6%_
## Footnote
* Number of moles (mol) = mass (g) ÷ relative formula mass *or* Mr
* % yield = 100 (actual mass ÷ theoretical mass)
89
# Question:
Nitrogen and hydrogen at high temperatures are converted to ammonia using the following reaction known as the Haber Process.
3H₂ + N₂ ⭢ 2NH₃
When 400 grams of H₂ are added to an excess amount of N₂, 104 grams of NH₃ are formed. Calculate the percent yield.
| **Triple only**
Step 1:
1. _Identify known and unknown substances._
Known: H₂ | Unknown: NH₃
2. _Mr of known and unknown substance._
Mr of H₂ = 2 | Mr of NH₃ = 17
3. _Moles of known substance._
Moles (mol) = 400g ÷ 2 = 200 mol
4. _Mole of unknown from ratio._
Ratio = 3:2 | Moles = 200 ÷ (3 ÷ 2) = 133.3 mol
5. _Mass of unknown._
[Theoretical] mass (g) = 133.3 mol × 17 = _2266g_
Step 2:
Percentage yield = 104g ÷ 2266g = 0.046 × 100 = **_4.6%_**
## Footnote
* Number of moles (mol) = mass (g) ÷ relative formula mass *or* Mr
* % yield = 100 (actual mass ÷ theoretical mass)
90
# Question:
If the typical yield is 86.78%, how much SO₂ should be expected if 4897 grams of ZnS are used? (2803 of SO₂)
ZnS + O₂ ⭢ ZnO + SO₂
| **Triple only**
Step 1:
1. _Identify known and unknown substances._
Known: ZnS | Unknown: SO₂
2. _Mr of known and unknown substance._
Mr of ZnS = 97 | Mr of SO₂ = 64
3. _Moles of known substance._
Moles (mol) = 4897g ÷ 97 = 50.5 mol
4. _Mole of unknown from ratio._
Ratio = 1:1 | Moles = 50.5 mol
5. _Mass of unknown._
[Theoretical] mass (g) = 50.5 mol × 64 = _3232g_
Step 2:
Actual mass = 86.78% × 3232g = **_2804.7g_**
## Footnote
* Number of moles (mol) = mass (g) ÷ relative formula mass *or* Mr
* Actual mass = % yield × theoretical mass
91
# Question:
A scientist reacted 130g of zinc with unlimited copper sulfate. They produced 25.4g of copper. Calculate the percentage yield.
Zn + CuSO₄ ⭢ Cu + ZnSO₄
| **Triple only**
Step 1:
1. _Identify the known and unknown._
Known: Zn | Unknown: Cu
2. _Ar of known and unknown._
Zn = 65 | Cu = 63.5
3. _Moles of known._
Moles = 130g ÷ 65 = 2 mol
4. _Moles of unknown from ratio._
Ratio = 1:1 | Moles = 2 mol
5. _Mass of unknown._
[Theoretical] mass = 2 mol × 63.5 = _127g_
Step 2:
Percentage yield = 25.4g ÷ 127g = 0.2 × 100 = *_20%_*
## Footnote
* Number of moles (mol) = mass (g) ÷ relative atomic mass *or* Ar
* % yield = actual mass ÷ theoretical mass
92
# Question:
A scientist reacted 27g of beryllium with unlimited hydrochloric acid. They produced 144g of beryllium chloride. Calculate the percentage yield.
Be + 2HCl ⭢ BeCl₂ + H₂
| **Triple only**
Step 1:
1. _Identify known and unknown._
Known: Be | Unknown: BeCl₂
2. _Mr and Ar of known and unknown._
Be = 9 | BeCl₂ = 80
3. _Moles of known._
Moles = 27g ÷ 9 = 3 mol
4. _Find moles of unknown from ratio._
Ratio = 1:1 | Moles = 3 mol
5. _Mass of unknown._
[Theoretical] mass = 3 mol × 80 = _240g_
Step 2:
Percentage yield = 144g ÷ 240g = 0.6 × 100 = *_60%_*
## Footnote
* Number of moles (mol) = mass (g) ÷ relative atomic mass *or* Ar
* % yield = actual mass ÷ theoretical mass
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What is atom economy
| **Triple only**
A measure of the amount of starting materials that end up as useful products.
94
Why is a high atom economy important in industry?
| **Triple only**
* By minimising the production of unwanted products, **money is saved**.
* By minimising the production of unwanted products, **sustainability increases** as resources are not wasted.
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What is the equation for calculating atom economy?
| **Triple only**
Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
96
What is the greatest atom economy we can have? Why?
| **Triple only**
100% because anything greater would mean we created atoms, which is not possible.
97
# Question:
Calculate the percentage atom economy for the production of silver iodide in this reaction.
The equation for the reaction is:
AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq)
Give your answer to 3 significant figures.
Relative formula masses:
AgNO₃ = 170 | NaI = 150 | AgI = 235 | NaNO₃ = 85
| **Triple only**
Atom economy = 235 ÷ 320 = (0.7340) × 100 = _73.4%_
| Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
98
# Question:
This question is about the extraction of metals.
Tungsten (W) is a metal. Tungsten is produced from tungsten oxide by reaction with hydrogen.
The equation for the reaction is:
WO₃ + 3H₂ → W + 3H₂O
Calculate the percentage atom economy when tungsten is produced in this reaction.
Relative formula masses (Mr):
WO₃ = 232 | H₂ = 2
| **Triple only**
Ar of W = 184
Atom economy = 184 ÷ (232 + (3 × 2)) = 0.773 × 100 = _77.3%_
| Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
99
How do chemists try to reduce waste when unwanted side products have been produced?
They use the side products in other reactions.
100
# Question:
This question is about reversible reactions and equilibrium.
Hydrogen is used to produce ammonia in the Haber process.
The hydrogen is made in two stages.
**Stage 1** is the reaction of methane and steam to produce carbon monoxide and hydrogen.
The equation for the reaction is:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
Calculate the atom economy for the formation of hydrogen in stage 1.
Relative atomic masses (Ar): H = 1 | C = 12 | O = 16
| **Triple only**
Atom economy = 3(1 × 2) ÷ 34 = 0.17647 × 100 = _17.7%_
| Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
101
# Question:
An equation for the reaction is:
NiO + C ⟶ Ni + CO
Calculate the percentage atom economy for the reaction to produce nickel.
Relative atomic masses (Ar): C = 12 | Ni = 59
Relative formula mass (Mr): NiO = 75
Give your answer to 3 significant figures.
| **Triple only**
Atom economy = 59 ÷ (75 + 12) = 0.6781 × 100 = _67.8%_
| Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
102
# Question:
(d) Look at the equations for the two reactions:
Reaction 1: CuCO₃(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l) + CO₂(g)
Reaction 2: CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l)
Calculate the percentage atom economy for Reaction 2 [Copper chloride].
Reactive formula masses: CuO = 79.5 | HCl = 36.5 | CuCl₂ = 134.5 | H₂O = 18
(e) The atom economy for Reaction 1 is 68.45%.
Compare the atom economies of the two reactions for making copper chloride.
Give a reason for the difference.
| **Triple only**
Part D:
Atom economy = 134.5 ÷ (79.5 + (2 × 36.5)) = 0.882 × 100 = _88.2%_
Part E:
Reaction 1 has a lower atom economy than reaction 2 because more side products were produced.
| Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
103
What is meant by concentration; what is the unit?
| **Triple only**
Concentration tells us the number of moles of a solute in a given volume of solution; mol/dm³.
104
What is the equation for calculating the concentration of a solution?
| **Triple only**
Concentration (mol/dm³) = number of moles ÷ volume (dm³)
105
# Question:
A solution has a concentration of 0.5 mol/dm³. Calculate the number of moles in 0.2 dm³.
| **Triple only**
Number of moles (mol) = 0.5 mol/dm³ × 0.2 dm³ = _0.01 moles_
| # of moles = concentration (mol/dm³) × volume (dm³)
106
# Question:
A solution has a concentration of 0.2 mol/dm³. Calculate the number of moles in 1.5 dm³.
| **Triple only**
Number of moles = 0.2 mol/dm³ × 1.5 dm³ = _0.3 moles_
| # of moles = concentration (mol/dm³) × volume (dm³)
107
# Question:
A solution of calcium chloride has a concentration of 0.4 mol/dm³. Calculate the mass of calcium chloride in 2 dm³.
Mr: CaCl₂ = 111
| **Triple only**
Step 1: Calculate the number of moles.
Number of moles = 0.4 mol/dm³ × 2 dm³ = _0.8 moles_
Step 2: Calculate the mass from moles and Mr.
Mass (g) = 0.8 moles × 111 = **_88.8g_** (1.d.p)
| # of moles = concentration (mol/dm³) × volume (dm³)
Mass (g) = # of moles (mol) × Relative formula mass *or* Mr
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# Question:
A solution of sodium nitrate has a concentration of 0.8 mol/dm³. Calculate the mass of sodium nitrate in 0.5 dm³.
Mr: NaNO₃ = 85
| **Triple only**
Step 1: Calculate the number of moles.
Number of moles = 0.8 mol/dm³ × 0.5 dm³ = _0.4 moles_
Step 2: Calculate the mass from moles and Mr.
Mass (g) = 0.4 moles × 85 = **_34g_**
| # of moles = concentration (mol/dm³) × volume (dm³)
Mass (g) = # of moles (mol) × Relative formula mass *or* Mr
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