LATEST EXAM PREP
Che1501
Memos + Marked
Questions and Answers
, (1) How many moles of hydrogen atoms does one mole of (NH4)2SO4 contain?
[1] 1 [2] 2 [3] 4 [4] 8 [5] 15
Each molecule of (NH₄)₂SO₄ has 8 Hydrogen atoms attached to it in total (4 x 2, from the
chemical formula), so, 1 mol of (NH₄)₂SO₄ will have 8 moles of Hydrogen.
Answer: [4]
(2) How many moles of NH3 are present in 100 g of NH3?
[1] 0.170 mol [2] 5.87 mol [3] 17.0 mol [4] 100 mol [5] 1703 mol
Molar mass of NH3 is 17.034 g.mol-1.
Number of moles, n = mass / molar mass, and n = 100 g / 17.034 g.mol-1 = 5.87 mol of
NH3 .
Answer: [2]
(3) What is the mass percentage of hydrogen in methane?
[1] 0.062 % [2] 0.251 % [3] 6.28 % [4] 12.6 % [5] 25.1 %
There is 4 moles of Hydrogen atoms in 1 mole of methane, CH4. The molecular mass of
1 hydrogen atom is 1.008 g per 1 mol of H atoms, or 4.032 g.mol-1 of 4 hydrogen atoms
in methane. Methane has molecular mass of 16.042 g per 1 mol of CH 4.
So, the mass percentage of hydrogen in methane = (4.032 g.mol-.042 g mol-1) x
100% = 25.1%.
Answer: [5]
(4) What is the empirical formula of a compound whose molecular formula is P4O10?
[1] PO [2] PO2 [3] P2O5 [4] P4O10 [5] P8O20
Note that the lowest whole-number relative amount of moles gives the empirical formula.
In P4O10, there are 4 moles of P atom and 10 moles of O atom. So the ratio of P : O is 4 :
10. However, the lowest ratio of P : O is 2 : 5 (we divided both by the common
denominator to get the lowest whole-number). Therefore, the empirical formula of a
compound whose molecular formula of P4O10 is P2O5.
Answer: [3]
(5) What is the empirical formula for a compound that contains 25.9 % nitrogen and 74.1 %
oxygen by mass?
[1] N2O4 [2] N4O6 [3] N2O5 [4] NO [5] NO5
Assume that you have a 100.0 g of sample of the compound. Accordingly, we have 25.9
g nitrogen and 74.1 g oxygen, giving 100.0 g sample of compound.
So 25.9 g N = (25.9 g N) x (1 mol N / 14.01 g N) = 1.85 mol N
And 74.1 g O = (74.1 g O) x (1 mol O / 16.00 g O) = 4.63 mol O.
So the ratio moles of N:O is 1.85:4.63
We divide by the lowest whole-number relative amount to get
1..85 : 4..85 = 1: 2.5
This ration gives the formula N1O2.5
Che1501
Memos + Marked
Questions and Answers
, (1) How many moles of hydrogen atoms does one mole of (NH4)2SO4 contain?
[1] 1 [2] 2 [3] 4 [4] 8 [5] 15
Each molecule of (NH₄)₂SO₄ has 8 Hydrogen atoms attached to it in total (4 x 2, from the
chemical formula), so, 1 mol of (NH₄)₂SO₄ will have 8 moles of Hydrogen.
Answer: [4]
(2) How many moles of NH3 are present in 100 g of NH3?
[1] 0.170 mol [2] 5.87 mol [3] 17.0 mol [4] 100 mol [5] 1703 mol
Molar mass of NH3 is 17.034 g.mol-1.
Number of moles, n = mass / molar mass, and n = 100 g / 17.034 g.mol-1 = 5.87 mol of
NH3 .
Answer: [2]
(3) What is the mass percentage of hydrogen in methane?
[1] 0.062 % [2] 0.251 % [3] 6.28 % [4] 12.6 % [5] 25.1 %
There is 4 moles of Hydrogen atoms in 1 mole of methane, CH4. The molecular mass of
1 hydrogen atom is 1.008 g per 1 mol of H atoms, or 4.032 g.mol-1 of 4 hydrogen atoms
in methane. Methane has molecular mass of 16.042 g per 1 mol of CH 4.
So, the mass percentage of hydrogen in methane = (4.032 g.mol-.042 g mol-1) x
100% = 25.1%.
Answer: [5]
(4) What is the empirical formula of a compound whose molecular formula is P4O10?
[1] PO [2] PO2 [3] P2O5 [4] P4O10 [5] P8O20
Note that the lowest whole-number relative amount of moles gives the empirical formula.
In P4O10, there are 4 moles of P atom and 10 moles of O atom. So the ratio of P : O is 4 :
10. However, the lowest ratio of P : O is 2 : 5 (we divided both by the common
denominator to get the lowest whole-number). Therefore, the empirical formula of a
compound whose molecular formula of P4O10 is P2O5.
Answer: [3]
(5) What is the empirical formula for a compound that contains 25.9 % nitrogen and 74.1 %
oxygen by mass?
[1] N2O4 [2] N4O6 [3] N2O5 [4] NO [5] NO5
Assume that you have a 100.0 g of sample of the compound. Accordingly, we have 25.9
g nitrogen and 74.1 g oxygen, giving 100.0 g sample of compound.
So 25.9 g N = (25.9 g N) x (1 mol N / 14.01 g N) = 1.85 mol N
And 74.1 g O = (74.1 g O) x (1 mol O / 16.00 g O) = 4.63 mol O.
So the ratio moles of N:O is 1.85:4.63
We divide by the lowest whole-number relative amount to get
1..85 : 4..85 = 1: 2.5
This ration gives the formula N1O2.5
