Exam I – Chapter 3 – Worksheet 3
Chapter 3 – Chemical Compounds
1. Be able to identify and draw the following from memory:
a. Functional groups presented in lecture notes
i. Haloalkanes, alcohols, ethers, aldehydes, ketones, carboxylic acids, esters, amines
ii. Alkanes, alkenes, alkynes
b. Some common names of inorganic molecular compounds
i. Ammonia, Hydrazine, Nitric oxide, Nitrous oxide, Phosphine
2. What would you name a molecular compound which consists of four oxygen and two nitrogen atoms?
How about three bromines and one phosphorous?
dinitrogen tetroxide
phosphorous tribromide
3. Predict the charges on ions of cesium, bromine, and nitrogen, and write symbols for these ions.
Cesium, Cs+ (alkali metal)
Bromine, Br- (halogen)
Nitrogen, N3- (Pnictogen)
Exam I – Chapter 3 – Worksheet 3
4. List five properties of molecular compounds, and five properties of ionic compounds
See lecture notes
5. Name the following ionic compounds and determine the charge on the metal:
CuI, , Li2CO3
Copper (I) Iodide, Cu+
Lithium Carbonate, Li+
𝑄 𝑄
6. Coulombs law: 𝐹⃑ = 𝑘 𝑑1 2 2
Assume Q1 and Q2 are +1 and -1 respectively and d=1unit
What happens to the force between two ions if Q2 is doubled? If d is halved?
If Q2 is doubled, the force doubles. If d is halved, the force increases by a factor of 4.
7. Determine the formula weights for each of the following ionic compounds, and find the percent
composition of oxygen in each one.
calcium acetate, aluminum(III) sulfate, iron(II) sulfate pentahydrate
-Ca(CH3COOH)2
molar mass of calcium acetate is 160.18 g/mol
4 ∗ 16
= .40
160.18
-Al2(SO4)3
molar mass of aluminum (III) sulfate is 342.15 g/mol
12 ∗ 16
= .56
342.15
-FeSO4*5H2O
molar mass of iron (II) sulfate pentahydrate is 241.98 g/mol
4 ∗ 16
= .59
241.98
Exam I – Chapter 3 – Worksheet 3
8. A major oil company has used an additive called MMT to boost the octane rating of its gasoline. What is
the empirical formula of MMT if it is 49.5% C, 3.2% H, 22.0% O, and 25.2% Mn.
49.5 𝑔 𝐶 1 𝑚𝑜𝑙 𝐶
∗
= 4.12 𝑚𝑜𝑙 𝐶
1
12.01 𝑔 𝐶
3.2 𝑔 𝐻 1 𝑚𝑜𝑙 𝐻
∗
= 3.17 𝑚𝑜𝑙 𝐻
1
1.01 𝑔 𝐻
22 𝑔 𝑂 1 𝑚𝑜𝑙 𝑂
∗
= 1.38 𝑚𝑜𝑙 𝑂
1
16 𝑔 𝑂
25.2 𝑔 𝑀𝑛 1 𝑚𝑜𝑙 𝑀𝑛
∗
= 0.46 𝑚𝑜𝑙 𝑀𝑛
1
54.94 𝑔 𝑀𝑛
The empirical formula is found by dividing through the molar amounts of each species by the species of
lowest quantity (here Mn). Thus, the empirical formula is C9H7MnO3
9. Acetylene is a colorless gas that is used as a fuel in welding torches, among other things. It is 92.26% C
and 7.74% H. It’s molar mass is 26.02 g/mol. Calculate the empirical and molecular formulas.
92.26 𝑔 𝐶 1 𝑚𝑜𝑙 𝐶
∗
= 7.68 𝑚𝑜𝑙 𝐶
1
12.01 𝑔 𝐶
7.74 𝑔 𝐻 1 𝑚𝑜𝑙 𝐻
∗
= 7.66 𝑚𝑜𝑙 𝐻
1
1.01 𝑔 𝐻
Thus, the empirical formula is C1H1.
The molecular weight of the CH fragment is 13.01 g/mol
Divide the molar mass by the molecular weight of the fragment
26.02
=2
13.01
Thus, the molecular formula for acetylene is C2H2