Center for Academic Services & Advising
September 9, 2016
Chemistry I – Chap 3
1. Complete the chart below:
Formula
Name
CSI Worksheet 1
Molecular Weight
HCl
Hydrochloric Acid
36.46 g/mol
CO
Carbon Monoxide
28.01 g/mol
Fe2O3
Iron(III) oxide
159.7 g/mol
MgCl2
Magnesium Chloride
95.21 g/mol
SiO2
Silicon Dioxide
60.09 g/mol
NH4NO3
Ammonium Nitrate
80.05 g/mol
NaHCO3
Sodium Bicarbonate
84.01 g/mol
2. The elemental mass percent composition of ascorbic acid (vitamin C) is 40.92% C, 4.58% H, and 54.50%
O. Determine the empirical formula for ascorbic acid.
Assume a basis of 100g of ascorbic acid, therefore:
C = 40.92 g, H = 4.58 g, O = 54.50 g
Convert all masses to moles:
1 mol C
= 3.41 mol C
12.01 g C
1 mol H
4.58 g H =
= 4.54 mol H
1.008 g H
1 mol O
54.50 g O =
= 3.41 mol O
16.00 g O
40.92 g C =
Write in the form of a chemical equation:
C3.41 H4.54 O3.41
Divide all subscripts by the lowest number (in this case, 3.41)
C1 H1.33 O1
Scale the formula to reach whole numbers in the subscripts
𝐂 𝟑 𝐇 𝟒 𝐎𝟑
COLORADO SCHOOL OF MINES
Catherine I Core Supplemental Instruction Facilitator | Chemistry I
Center for Academic Services & Advising
3. Write a balanced chemical equation for the following reaction:
Solid iron(III) oxide reacts with hydrogen gas to form solid iron and liquid water.
Determine the unbalanced formula first:
Fe2 O3 (s) + H2 (g) → Fe(s) + H2 O(l)
Next, balance the equation:
 There are 2 iron atoms in the Fe2O3 (iron(III) oxide), so there must be 2 iron atoms on
the products side of the equation.
 There are 3 oxygen atoms in the Fe2O3 (iron(III) oxide), so there must be 3 oxygen
atoms on the products side of the equation.
 There are 2 hydrogen atoms on the reactants side, and since liquid water is already
scaled up by 3 to account for the oxygen, there must be 3 hydrogen molecules on the
reactants side.
Fe2 O3 (s) + 𝟑 H2 (g) → 𝟐 Fe(s) + 𝟑 H2 O(l)
4. Fructose is a common sugar found in fruit. Elemental analysis of fructose gives the following mass percent
composition: 40.00% C, 6.72% H, and 53.28% O. The molar mass of fructose is 180.16 g/mol. Find the
molecular formula for fructose.
Assume a basis of 100g of fructose:
C = 40.00 g, H = 6.72 g, O = 53.28 g
Convert all masses to moles:
1 mol C
= 3.33 mol C
12.01 g C
1 mol H
6.72 g H =
= 6.67 mol H
1.008 g H
1 mol O
53.28 g O =
= 3.33 mol O
16.00 g O
Write in the form of a chemical equation:
C3.33 H6.67 O3.33
Divide all subscripts by the lowest number (in this case, 3.33)
C1 H2 O1
Determine the “molecular weight” of this empirical formula:
𝑔
𝑔
𝑔
1 𝐶 ∗ 12.01
+ 2 𝐻 ∗ 1.008
+ 1 𝑂 ∗ 16.00
= 30.03 𝑔/𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
Take the molar mass of fructose and divide by the calculated molar mass:
𝑔
180.16
𝑚𝑜𝑙 = 6.00
𝑔
30.03
𝑚𝑜𝑙
Therefore, the molecular formula for fructose is:
6 ∗ (C1 H2 O1 ) = 𝐂𝟔 𝐇𝟏𝟐 𝐎𝟔
40.00 g C =
COLORADO SCHOOL OF MINES
Catherine I Core Supplemental Instruction Facilitator | Chemistry I