Review Chapter 10
1. How many particles are present in a mole? This is known as ______number.
2. What does the atomic mass represent?
3. How many molecules are in 2.6 moles of CO?
4. How many atoms of hydrogen are in 2.50 moles Ammonia, NH3 ?
5. How many moles are in 22.0 grams of copper metal?
6. What is the molar mass of aluminum oxide?
7. What would the mass be of 3.7 moles of Fe2O3?
% composition from a Formula
8. Determine the percent composition for each of the elements in magnesium nitrate
% composition from a Mass Data
9. What is the % composition of this compound – a 45.0 gram sample containing 35.1 grams of iron and 9.90 grams of oxygen?
10. What is the % composition of this compound - 75.0 gram sample containing 20.5 grams of carbon and 54.5 grams of oxygen?
Determining the Empirical Formula of a Compound
11. Determine the empirical formula for a compound with the following elemental composition: 40.00% C, 6.72% H, 53.29% O.
12. In a 100 g sample a compound contains 57.1 g carbon, 38.0 g sulfur and 4.9 g of hydrogen. Determine the empirical formula.
Determining the Molecular Formula of a Compound
13. Imagine that the molar mass of the compound found in Question 11 was 180g/mol. What would its molecular formula be?
14. Imagine that the molar mass of the compound found in Question 12 was 168 g/mol. What would its molecular formula be?
ANSWER KEY
1. 6.02x1023 / Avogandro’s
2. The mass of one mole of the element.
3. 1.57 x1024
4
5. 22.0/63.55 = 0.346 moles
6. (2 x 26.98) + (3x16.00) = 101.96 g/mol
7. 3.7 x (2 x 55.85 + 3 x 16.00) = 590.89 g
8 M of Mg(NO3)2 : 24.30 g/mol + 2(14.01 g/mol) + 6(16.0 g/mol) = 148.32 g/mol
% Mg = 24.30 g/mol x 100 = 16.4 %
148.32 g/ mol
% N = 2(14.01 g/mol) x 100 = 18.9 %
148.32 g/ mol
% O = 6 (16.00 g/mol) x100 = 64.7%
9. % Fe = 35.1 g x 100 = 78.0 %
45.0g
. % O = 9.90 g x 100 = 22.0 %
45.0g
10. % C = 20.5 g x 100 = 27.3.0 %
75.0g
. % O = 54.5 g x 100 = 72.7 %
75.0g
11.
divide through each of the mole quantities by which ever mole quantity is the smallest number of moles. In this example, the smallest mole quantity is either the moles of carbon or moles of oxygen (3.331 mol):
The ratio of C:H:O has been found to be 1:2:1, thus the empirical formula is: CH2O.
12.
The ratio of C:H:S has been found to be 4:4:1, thus the empirical formula is: C4H4S
13. The formula weight of the empirical formula is 30 g/mol. Divide the molecular weight by the empirical formula weight to find a multiple:
The molecular formula is a multiple of 6 times the empirical formula:
C(1 x 6) H(2 x 6) O(1 x 6) which becomes C6H12O6
14. The molar mass of the empirical formula is 84 g/mol. Since the molecular weight of the actual compound is 168 g/mol, and is double the molar mass of the empirical formula, the molecular formula must be twice the empirical formula:
C(4 x 2) H(4 x 2) S(1 x 2) which becomes C8H8S2