Week 3 - Stoichiometry

Question 1

Balance the equation AgNO3 + H2S → Ag2S + HNO3

Answer 1

There are two Ag on the right but only one on the left. Put 2 in front of AgNO3.

2 AgNO3 + H2S → Ag2S + HNO3

There are now two N on the left and one on the right. Put 2 in front of HNO3.

2 AgNO3 + H2S → Ag2S + 2 HNO3

Now check - there are two Ag on each side, two N on each side, six O on each side (2 × 3), two H on each side, and two H on each side.

Question 2

Balance the equation C4H10 + O2 → CO2 + H2O

Answer 2

There are four C on the left but only one on the right. Put 4 in front of CO2.

C4H10 + O2 → 4 CO2 + H2O

There are ten H on the left and two on the right. Put 5 in front of H2O.

C4H10 + O2 → 4 CO2 + 5 H2O

There are two O on the left and 13 on the right (8 + 5). To have 13 O on the left, you would need 6.5 O2. However, we need to use whole numbers. Let’s double everything:

2 C4H10 + O2 → 8 CO2 + 10 H2O

Now we have 26 O on the right. Put 13 in front of the O2.

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Now check - we have 8 C, 20 H, and 26 O on each side.

Question 3

What are the symbols used in balanced equations?

Answer 3

Symbol Meaning
(s) Solid state (written after substance)
(l) Liquid state (written after substance)
(g) Gaseous state (written after substance)
(aq) Aqueous state (when substance is dissolved in water)
Δ Heat is added (when written above or below the arrow)

Question 4

Symbols and meanings of chemical equations

Answer 4

Symbol Meaning
(s) Solid state (written after substance)
(l) Liquid state (written after substance)
(g) Gaseous state (written after substance)
(aq) Aqueous state (when substance is dissolved in water)
Δ Heat is added (when written above or below the arrow)

Question 5

What is a decomposition reaction?

Answer 5

A single substance is decomposed, or broken down, to give two or more different substances.

2 HgO(s) → 2 Hg(l) + O2(g)
CaCO3(s) → CaO(s) + CO2(g)
2 KClO3(s) → 2 KCl(s) + 3O2(g)

Question 6

What is a single-displacement reaction?

Answer 6

One element reacts with a compound to replace one of the elements of that compound, yielding a different element and a different compound.

Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq)
2 Al(s) + 3 H2SO4(aq) → 3 H2(g) + Al2(SO4)3(aq)

Question 7

What is a double-displacement reaction

Answer 7

Two compounds exchange partners with each other to produce two different compounds.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
CuO(s) + 2 HNO3(aq) → Cu(NO3)2(aq) + H2O(l)

Question 8

Equation for percentage composition (formula) of molecules

Answer 8

% element = (number of atoms of the element) x (atomic mass of element) / formula mass of compound x 100

Question 9

Avogadro’s number

Answer 9

A mole used to be defined as the amount of matter that contains as many objects as the number of atoms in exactly 12 g of pure 12C. 12 g of has 6.022 × 1023 of atoms.

In 2017, a new definition of mole was proposed, with IUPAC Recommendations, to state:

      “One mole contains exactly 6.022 140 76 × 1023 elementary entities. This number is the fixed             numerical value of the Avogadro constant, NA, when expressed in mol−1, and is called the                  Avogadro number (Pure and Applied Chemistry 2018, 90(1), p175–180).”

Question 10

Molar Mass

Answer 10

The atomic mass of an element in grams contains Avogadro’s number of atoms and is defined as the molar mass of that element.

To determine molar mass, we change the units of the atomic mass (found in the periodic table) from atomic mass unit (amu) to grams.

For example, sulfur has atomic mass of 32.07 amu (for one sulfur atom), so 1 mol (6.022×1023 sulfur atoms) of sulfur has a molar mass of 32.07 g.

Question 11

formula for Mole

Answer 11

Mole = Mass (g) / Molar Mass

Question 12

How many magnesium atoms are contained in 5.00g Mg?

Answer 12

Molar mass of Mg = 24.31 g/mol, therefore,
Moles = Mass / Molar Mass = 5.00 g / 24.31 g/mol = 0.206 mol

Because 1 mole contains 6.022 × 1023 items (Mg atoms in this case), 0.206 mol has 0.206 mol × 6.022 × 1023 = 1.24 × 1023 Mg atoms

Question 13

Define Stoichiometry

Answer 13

Measuring chemicals that go into, and come out of, and given reaction

Question 14

How to find mass (equation)

Answer 14

Mass (g) = moles (mol) x molar mass (g/mol)

Question 15

How many molecules in 40.03 mg of ammonium nitrate (NH4)NO3?

Answer 15

Referring to the above equations we first need to calculate the molar mass of ammonium nitrate, then calculate the number of moles this equated to and finally we can determine the number of molecules in that amount.

1

1
Calculate the molar mass of ammonium nitrate

2 N = 2 x 14.01 = 28.02 g/mol

4 H = 4 x 1.01 = 4.04 g/mol

3 O = 3 x 16.00 = 48.00 g/mol

Molar mass = = 80.06 g/mol

2

2
Calculate the number of moles of ammonium nitrate

Moles = 40.03 mg / 80.06 g/mol

Moles = 5.000 x 10-4 moles

3

3
Calculate the number of molecules of ammonium nitrate

Number of molecules = moles x Avogadro’s Number

Number of molecules = 5.000 x 10-4 x 6.022 x 1023

Number of molecules = 30.11 x 1019 molecules of ammonium nitrate

Question 16

Difference between empirical and molecular formula

Answer 16

The empirical formula, or simplest formula, gives the smallest whole-number ratio of atoms present in a compound.
The molecular formula is the true formula, representing the total number of atoms or each element present in one molecule of a compound.
For example, acetylene (C2H2) is a common gas used in welding; Benzene (C6H6) is an important solvent used in synthesis of styrene and nylon. Both contain 92.3% C and 7.7% H. The smallest ratio of C and H for both compounds is 1:1. They have the same empirical formula but different molecular formulae.

Question 17

What is a limiting reactant

Answer 17

For example, using 80 wheels and 50 frames to assemble bicycles, the number of bicycles that can be built from these parts is determined by the “limiting reactant” (the wheels in this case). There are enough frames for 50 bicycles, but only enough wheels for 40 bicycles. Once those wheels have been used, you can’t make any more bicycles.

Question 18

Theoretical yield and formula

Answer 18

The quantities of the products we have been calculating from chemical equations represent the maximum yield (100%), called the theoretical yield, of product according to the reaction represented by the equation.

percent yield = actual yield/theoretical yield *100

Question 19

(Molarity)

What is the molarity of a solution made by dissolving 5.85 g of sodium chloride (NaCl) in water to make 250. mL of solution?

Answer 19

Molarity (mol/L) = moles / volume (L)

    Molar mass of NaCl= 58.45 g/mol

    Moles of NaCl =  5.85 g / 58.45 g/mol = 0.100 mol

    Molarity = 0.100 mol / 0.250 L = 0.400 M

Question 20

(Molarity)

What is the molarity of a solution made by dissolving 2.00 g of potassium chlorate (KClO3) in enough water to make 150. mL of solution?

Answer 20

Molarity (mol/L) = moles / volume (L)

    Molar mass of KClO3 = 122.6 g/mol

    Moles of KClO3 =  2.00 g / 122.5 g/mol = 0.0163 mol

    Molarity = 0.0163 mol / 0.150 L = 0.109 M

Question 21

Dilution

Answer 21

As only water is added in the dilution process, the moles of solute before dilution are the same as the moles of solute after dilution. We can express this in the following equation:

MiVi = moles = MfVf or simply as MiVi = MfVf

Where the subscript i represents the initial and the subscript f the final molarity and volumes of a solution.

This is sometimes written as C1V1 = C2V2

Question 22

Dilution question

What mass of lead chloride [PbCl2] could be obtained when 50 mL of a 0.200 M lead nitrate solution [Pb(NO3)2] is reacted with 250 mL of 0.100 M hydrochloric acid [HCl]?

Answer 22

Step 1 : Start with the balanced equation:

    Pb(NO3)2 (aq) + 2 HCl (aq) → PbCl2 (s) + 2 HNO3 (aq)

Step 2 : Calculate the number of moles of lead nitrate and hydrochloric acid

Moles = Molarity (mol/L) x volume (L)

        Lead Nitrate:    0.200 mol/L x 0.050 L = 0.01 mol

        Hydrochloric acid:    0.100 mol/L x 0.250 L = 0.025 mol

Step 3 : Determine which is the limiting reagent by dividing by the reaction coefficient

        Lead nitrate = 0.01 mol /1 = 0.01 mol

        Hydrochloric acid = 0.025 mol/2 = 0.0125 mol

        Lead nitrate is therefore the limiting reagent

Step 4 : Calculate the mass lead chloride which could be obtained.

       0.01 moles lead nitrate would react to give 0.01 moles lead chloride (1:1 ratio)

   Mass (g) = moles x molar mass (g/mol)    

       Mass lead chloride (g) = 0.01 mol x 278.1 g/mol = 2.78 g