Name: __________________________________________________
Homework 2
1.) Name the following ionic compounds.
a. MgO ____magnesium oxide__________________
b. Al2S3 ____aluminum sulfide_____________
c. Na2SO4 ____sodium sulfate____
2.) Name the following covalent compounds:
a. NO2 ___nitrogen dioxide_____________________
b. SO2 _____sulfur dioxide____________________
c. PCl5 _____phosphorus pentachloride___________
d. PCl3 _____phosphorus trichloride___________
3.) Vanillic Acid, an oxidation product of the artificial flavoring vanillin, has the following elemental
analysis: C 57.14%, H 4.80%, O 38.06%. Determine the empirical formula.
1 mole C
= 4.757702 moles C/ 2.37875 = 2 C
12.01 grams C
57.14 grams C x
4.80 grams H x
1 mole H
= 4.752475 moles H/ 2.37875 = 1.997 = 2 H
1.01 grams H
38.06 grams O x
1 mole O
= 2.37875 moles O/ 2.37875 = 1 O
16.00 grams O
C2H2O - empirical
4.) A 3.00 mg of aspirin was analyzed by combustion. Aspirin is known to contain 3 elements:
carbon, hydrogen, and oxygen (formula: CxHyOz). The results of the combustion analysis found
6.60 mg of CO2 and 1.20 mg of H2O. Determine the empirical formula
CxHyOz →
3.00 mg
6.60 mg CO2 x
1.20 mg H2O x
1g
1000 mg
1g
1000 mg
CO2 +
6.60 mg
H2O
1.20 mg
x
12.01 grams C 1000 mg C
1 mole CO 2
1mole C
x
x
x
= 1.80mg C
44.01 g CO 2 1 mole CO 2
1 mole C
1g C
x
1.01 grams H 1000 mg H
1 mole H 2 O
2 mole h
= 0.135 mg H
x
x
x
18.02 g H 2 O 1 mole H 2 O
1 mole H
1g H
mg C + mg H + mg O = total mg of sample
1.80 mg C + 0.135 mg H + x = 3.00 mg total
x = mg O = 1.07 mg
2.25 x 4 = 9
1.99 = 2 x 4 = 8
=1x4=4
Moles O are the smallest – divide everyone by moles O!
Since C is not a whole number we must turn all species into whole numbers
Empirical formula = C9H8O4
5.) Potassium iodide is used as a dietary supplement to prevent the iodine deficiency disease, goiter,
and is prepared by the reaction of hydroiodic acid and potassium hydrogen carbonate. Water,
carbon dioxide and potassium iodide are the products of this reaction. In reacting 245 grams of
hydroiodic acid with 116 grams of potassium hydrogen carbonate, how many grams of
potassium iodide are produced? Which reactant is in excess? Circle the limiting reactant.
First, you need the balanced equation: 1HI + 1KHCO3 → 1H2O + 1CO2 + 1KI
Second: Since information was given to us about BOTH reactants, you need to determine the
LR. The LR is NOT the species that is present in the smallest amount initially – it is the
species that PRODUCES the smallest amount of product. We cannot simply calculate the
number of moles of species and say, ohh the smallest one is always the LR. It just isn’t true!
116 grams KHCO3 x
166.0 grams KI
1 mole KHCO 3
1 moleKI
= 192 grams KI
x
x
100.12 grams HKHCO 3 1 mole KHCO 3
1 mole KI
(theor)
245 grams HI x
1 mole HI
1 moleKI 166.0 grams KI
= 318 grams of KI (theor)
x
x
127.9 grams HI 1 mole HI
1 mole KI
So, the KHCO3 gives us one number, and the HI gives us another amount of KI that could be
made. What the smaller amount of KI tells you is that after 192 grams of KI has been made,
there is no more KHCO3 – it is all gone. If there is no more KHCO3, then the HI cannot be
converted into the products at all! So KHCO3 is the limiting reactant and 192 grams is the
theoretical yield!