# Combined Chemistry - C3 Quantitative Chemistry Flashcards

What does the law of the conservation of mass state?

No atoms are lost or made during a chemical reaction

A piece of magnesium is added to acid. What would happen to the mass reading on the balance and why?

It would **stay the same** because there is a lid on the conical flask so no gas can escape.

Count the number of elements in the following compounds:

a) CaO

b) Al_{2}O_{3}

c) H_{2}SO_{4}

a) CaO - 2 elements (Ca and O)

b) Al_{2}O_{3 }- 2 elements (Al and O)

c) H_{2}SO_{4} - 3 elements (H, S and O)

Count the number of atoms in the following compounds:

a) CaO

b) Al_{2}O_{3}

c) H_{2}SO_{4}

a) CaO - 2 atoms (1 x Ca and 1 x O)

b) Al_{2}O_{3} - 5 atoms (2 x Al and 3 x O)

c) H_{2}SO_{4} - 7 atoms (2 x H, 1 x S and 4 x O)

Calculate the relative formula mass (Mr) of each of the compounds (a, b & c) below.

AR: Mg = 24; Cl = 35.5; Ca = 40; O = 16; Al = 27

a) CaO

b) MgCl_{2}

c) Al_{2}O_{3}

*this is just an example, you need to be able to do this for any compound.

a) CaO: 40 + 16 = **56**

b) MgCl_{2}: 24 + (35.5x2) = **95**

c) Al_{2}O_{3}: (27x 2) + (16 x 3) = **102**

A piece of magnesium is added to acid. What would happen to the mass reading on the balance and why?

It would **decrease** because there is no lid on the conical flask so gas can escape and therefore mass is ’lost’ from the flask.

Calculate the uncertainty in the following sets of results

a) 1, 3, 6, 7, 10

b) 22, 25, 26, 25, 24

*this is just an example, you need to be able to do this for any set of data given.

Use the equation: uncertainty = range / 2

a) 1, 3, 6, 7, 10

Range = 10 – 1 = 9. Uncertainty = 9/2

**Uncertainty = ± 4.5**

b) 22, 25, 26, 25, 24

Range = 26 – 22 = 4. Uncertainty = 4/2

**Uncertainty = ± 2**

Calculate the percentage mass of hydrogen in:

Ar: N = 14 ; H = 1 ; S = 32 ; O = 16.

a) NH_{3}

b) H_{2}SO_{4}

Give your answer to the nearest whole number.

*this is just an example, you need to be able to do this for any compound when given the appropriate information.

% mass = (Ar x number of atoms of the element) / (Mr of compound) x 100

a) NH_{3} : (3x1) / 17 = 0.176

0. 176 x 100 = **18 %**

b) H_{2}SO_{4} : (2x1) / 98

0. 02 x 100 = **2 %**

**Higher Q**. How many molecules, atoms or ions are in one mole of a substance?

6.02 x 10^{23}

Higher Q**. How many moles are in the following?**

a) 46 g of CaO?

b) 50 g of O_{2}?

Number of moles = (mass in g)/Mr

a) Mr of CaO = 40 + 16 = 56.

Number of moles = 46 g /56 = **0.82 moles**

b) Mr of O_{2} =16

Number of moles = 50g / 32 = **1.6 moles**

**Higher Q**. Describe the equations below in words in terms of the number of moles of each substance reacting.

1) 2Mg + O_{2} → 2MgO

2) H_{2}SO_{4} + 2NaOH → Na_{2}SO_{4} + 2H_{2}O

1) 2 moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide.

2) One mole of sulfuric acid reacts with two moles of sodium hydroxide to produce one mole of sodium sulfate and two moles of water.

**Higher Q**. Calculate the mass of magnesium needed to produce 4.6 tonnes of magnesium oxide.

2Mg + O_{2} → 2MgO

*this is just an example, you need to be able to do this for any equation when given a mass of a reactant or product.

Step 1 – Make sure the equation is balanced - This equation is already balanced.

Step 2 – Calculate the Mr of 2Mg and the Mr of 2MgO.

Mr of 2Mg = 48, Mr of 2MgO =80

Step 3 – You now know 48 tonnes of magnesium would produce 80 tonnes of magnesium oxide.

Step 4 – Find how much magnesium you would need to produce 1 tonne of magnesium oxide: (48/80) magnesium and (80/80) magnesium oxide.

Therefore 0.6 tonnes of magnesium would produce 1 tonne of magnesium oxide.

Step 5 – Multiply both sides by 4.6 to work out how much magnesium would be needed to make 4.6 tonnes of magnesium oxide.

(0.6 x 4.6) = **2.76 tonnes**

**Higher Q**. Calculate the mass of CO_{2} produced when 50 g of CaCO_{3} is thermally decomposed.

CaCO_{3} → CaO + CO_{2}

*this is just an example, you need to be able to do this for any equation when given a mass of a reactant or product.

Step 1 – Make sure the equation is balanced - This equation is already balanced.

Step 2 – Calculate the Mr of CaCO_{3} and the Mr of CO_{2}.

Mr of CaCO_{3} = 100, Mr of CO_{2} = 44

Step 3 – You now know 100 g of CaCO_{3} would produce 44 g of CO_{2}.

Step 4 – Find what 1 g of CaCO_{3} would produce.

(100/100) CaCO_{3} and 44/100 CO_{2}

1g of CaCO_{3} would produce 0.44 g of CO_{2}

Step 5 – Multiply both sides by 50 to work out what 50 g of CaCO_{3} would make.

50 g of CaCO_{3} would make (0.44 x 50g) of CO_{2}

50 g of CaCO_{3} would make **22g of CO _{2}**

**Higher Q**. Write a balanced symbol equation for the following reaction:

48 g of magnesium (Mg) reacts with 32 g oxygen (O_{2}) to produce 80 g of magnesium oxide (MgO).

*this is just an example, you need to be able to do this for any equation when given the appropriate information.

Find ratio of moles in equation by finding number of moles of each substance using the equation:

Number of moles=(mass in g)/Mr

Mg: 48/24 = 2 ; O_{2}: 32/32 =1 ; MgO: 80/40 = 2

Ratio = 2Mg: 1O2 : 2MgO

Balanced equation: **2Mg + O _{2} → 2MgO**

**Higher Q**. What is a ‘limiting reactant’?

The first reactant that gets used up in a reaction and causes the reaction to stop.