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.

CalciumOxygen
Mass0.40.16
Molar mass4016
Number of moles0.4/40 = 0.010.16/16 = 0.01
Molar ratio11

So the empirical formula is CaO.

b.

CarbonHydrogen
Mass8020
Molar mass121
Number of moles80/12 = 6.6720/1 = 20
Molar ratio6.67/6.67 = 120/6.67 = 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.

ElementAr
H1
C12
N14
O16
Na23
Mg24
Si28
S32
Cl35.5
Fe56

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

SiliconOxygen
Mass0.280.32
Molar mass2832
Number of moles0.28/28 = 0.010.32/16 = 0.02
Molar ratio0.01/0.01 = 10.02/0.01 = 2

Empirical formula = SiO2

Sydney Chako

Mathematics, Chemistry and Physics teacher at Sytech Learning Academy. From Junior Secondary School to Tertiary Level Engineering Mathematics and Engineering Science.

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