# Stoichiometry Questions and Answers(Page 1)

Contents

## Stoichiometry Questions and Answers (Page 1)

below.
[H = 1; C = 12; N = 14; O = 16; Ne = 20; Na = 23;
Mg = 24; S = 32; K = 39; Fe = 56; Cu = 63.5; Zn = 65.
]

One mole of any gas at rtp occupies 28 dm3.

## Question 1

Calculate the number of moles in:
a. 2 g of neon atoms
b. 4 g of magnesium atoms
c. 24g of carbon atoms.

## Solution 1

a. Number of moles = mass ÷ molar mass
= 2/20
= 0.1 moles

b. Number of moles = mass ÷ molar mass
= 4/24
= 0.167 moles

c. Number of moles = mass ÷ molar mass
= 24/12
= 2 moles

## Question 2

Calculate the mass of:
a. 0.1 mole of oxygen molecules
b. 5 moles of sulfur atoms
c. 0.25 mole of sodium atoms

## Solution 2

a. each oxygen molecule contains 2 atoms of oxygen. So since the Mr of an oxygen atom is 16, the molar mass of oxygen molecules is 2 × 16 = 32g
Mass = Number of moles × molar mass
= 0.1 × 32
= 3.2 g

b. Mass = Number of moles × molar mass
= 5 × 32
= 160g

c. Mass = Number of moles × molar mass
= 0.25 × 23
= 5.75g

## Question 3

Use the following values of Ar to answer the questions below. [H = 1; C = 12; O = 16; Ca = 40.] a. Determine the empirical formula of an oxide of calcium formed when 0.4g of calcium reacts with 0.16g of oxygen.
b. Determine the empirical formula of an organic hydrocarbon compound which contains 80% by mass of carbon and 20% by mass of hydrogen. If the Mr of the compound is 30, what is its molecular formula?

## Solution 3

a.

Calcium Oxygen
Mass 0.4 0.16
Molar mass 40 16
Number of moles 0.4/40 = 0.01 0.16/16 = 0.01
Molar ratio 1 1

So the empirical formula is CaO.

b.

Carbon Hydrogen
Mass 80 20
Molar mass 12 1
Number of moles 80/12 = 6.67 20/1 = 20
Molar ratio 6.67/6.67 = 1 20/6.67 = 3

So the empirical formula is CH3

To find molecular formula we multiply empirical formula by number of formula units.

Formula units = Molecular mass ÷ Empirical mass
= 30 ÷ (12 + 3)
= 30 ÷ 15
= 2

Molecular mass = 2(CH3)
= C2H6

## Question 4

Use the data in the table below to answer the questions which follow.

Element Ar
H 1
C 12
N 14
O 16
Na 23
Mg 24
Si 28
S 32
Cl 35.5
Fe 56

Calculate the mass of:
a. 1 mole of:
(i) chlorine molecules
(ii) iron(iii) oxide.
b. 0.5 mole of magnesium nitrate

## Solution 4

a.(i) 35.5g
(ii) iron(iii) oxide is Fe2O3
mass of 1 mole = (56 × 2 + 16 × 3) = 160g

b.(i) molar mass of Mg(NO3)2
= 24 + 14 × 2 + 16 × 6
= 148g
therefore mass of 0.5 moles = 0.5 × 148
= 74g

## Question 5

NB the molar gas molar varies according to local conditions, as such, for cambridge exams its 24 dm3 and for zimsec exams its 28 dm3.
Calculate the volume occupied, at rtp, by the following gases.
a. 12.5 moles of sulfur dioxide gas.
b. 0.15 mole of nitrogen gas.

## Solution 5

a. Volume = number of moles × molar gas volume
= 12.5 × 28 dm3
= 350 dm3

b. Volume = number of moles × molar gas volume
= 0.15 × 28 dm3
= 4.2 dm3

## Question 6

Calculate the number of moles of gas present in the following:
a. 36cm3 of sulfur dioxide
b. 144 dm3 of hydrogen sulfide.

## Solution 6

a. since the volume is given in cm3 we can as well first convert the molar gas volume to cm3. 28 dm3 = 28 000 cm3.
Number of moles = volume ÷ molar volume
= 36/28 000
= 0.0013 moles

b. Number of moles = volume ÷ molar volume
= 144/28
= 5.14 moles

## Question 7

Use the following experimental information to determine the empirical formula of an oxide of silicon.

Mass of crucible = 18.20 g
Mass of crucible + silicon = 18.48 g
Mass of crucible + oxide of silicon = 18.80 g

## Solution 7

To find the empirical formula of a compound we need the masses of the elements that make up the compound.

Mass of silicon = 18.48 – 18.20 = 0.28g
Mass of oxygen = 18.80 – 18.48 = 0.32g

Silicon Oxygen
Mass 0.28 0.32
Molar mass 28 32
Number of moles 0.28/28 = 0.01 0.32/16 = 0.02
Molar ratio 0.01/0.01 = 1 0.02/0.01 = 2

Empirical formula = SiO2

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