Chemistry 139
Clark College
Name: _____________________________
Homework 5: formulas and molar masses, moles and calculations
Show your work, use units and sig figs, where appropriate.
1. Write the ions that each species would form, pair them up to make a neutral compound and make
sure to clearly indicate the charges on the ions and remember that the final formula must be
neutral: (you might need your text or the nomenclature handout on the class website!)
a) K and P ___ K+1 + P-3  K3P ________________
b) Al and N ___ Al+3 + N-3  Al3N3  AlN ___________
c) Al and nitrate ___ Al+3 + NO3-1  Al(NO3)3____________
d) Ammonium and Cl ___ NH4+1 + Cl-1  NH4Cl ______________
2. Write the formula for the following compounds
a) Dinitrogen trioxide ________ N2O3_____________________
b) Cesium sulfate ________ Cs2SO4_______________________
c) Vanadium (II) carbonate ____ VCO3_____________
3. Give the name for the following compounds.
a) Al2O3 _____ aluminum oxide_____________
b) P2O3 _____ diphosphorus trioxide ___
c) Cr(MnO4)3 __ chromium (III) permanganate ______
d) Ca(OH)2 ___ calcium hydroxide _________
e) P4O10 ___ tetraphosphorus decoxide ______
f) BrCl ____ bromine monochloride _____
g) Ca(NO3)2 __ calcium nitrate ___
h) KMnO4 ___ potassium permanganate ____
i) (NH4)2SO4 ____ ammonium sulfate ________
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Chemistry 139
Clark College
use these as models for your homework problems. You should SHOW ALL WORK – all steps
in your conversions!
10.33 grams Na3PO4 x
1 mole Na 3 PO 4
= 0.06301 moles Na3PO4
163.94 g Na 3 PO 4
10.33 grams Na3PO4 x
1 mole Na 3 PO 4
3 moles Na 1 = 0.1890 moles Na+1 ions
x
163.94 g Na 3 PO 4 1 mole Na 3 PO 4
10.33 rams Na3PO4 x
1 mole Na 3 PO 4
3 moles Na 1
6.022 x 10 23 Na 1 ions = 1.139 x 1023 Na+1 ions
x
x
163.94 g Na 3 PO 4 1 mole Na 3 PO 4
1 mole Na 1
4. How many formula units are in 10.35 moles of magnesium phosphate? (HINT: write the
formula first)
Magnesium phosphate = Mg+2 PO4-3 → Mg3(PO4)2
6.022x10 23 f.u. Mg 3 (PO 4 ) 2
10.35 moles Mg3(PO4)2 x
= 6.233
1mole Mg 3 (PO 4 ) 2
x 1024 f.u. of Mg3(PO4)2
5. How many calcium ions are there in 0.625 moles of calcium nitride? (HINT: write the
formula first)
Calcium nitride = Ca+2 N-3 → Ca3N2
0.625 moles Ca3N2 x
3 mole Ca 2 6.022x10 23 Ca 2 ions
x
= 1.13
1mole Ca 3 N 2
1 mole Ca 2
x 1024 Ca+2 ions
6. How many carbon atoms are found in 2.35 grams of acetic acid, HC2H3O2 (aq)?
mm = 2 C x (12.01 g/mole) + 4 H (1.01 gram/mole) + 2 O (16.00 gram/mole) = 60.06 gram/mole
2.35 grams HC2H3O2 x
1 mole HC 2 H 3 O 2
2 mole C
6.022x10 23 C atoms
= 4.71 x 1022
x
x
60.06 grams HC 2 H 3 O 2 1mole HC 2 H 3 O 2
1 mole C
C atoms
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Chemistry 139
Clark College
7. A reaction requires 1.10 mol of vanadium (IV) chloride. How many grams of vanadium
(IV) chloride should you weigh out? (Hint: write that formula first!)
V(IV) = V+4 Cl-1 → VCl4
V : 1 x 50.94 = 50.94 + Cl: 4 x 35.45 = 141.8 = 192.7 grams/mole
1.10 moles VCl4 x
192.7 grams VCl 4
= 211.97 = 212
1 mole VCl 4
grams VCl4
8. How many atoms of nitrogen are in 125.9 g of manganese (II) nitrate? (HINT: write that
formula first!)
Mn+2 NO3-1 → Mn(NO3)2
(1 Mn x 54.94) + (2N x 14.01) + (6O x 16.00) = 178.96 grams/mole
125.9 grams Mn(NO3)2 x
1 mole Mn(NO 3 ) 2
2 mole N
6.022x10 23 N atoms =
x
x
178.96 grams Mn(NO 3 ) 2 1 mole Mn(NO 3 ) 2
1 mole N
8.476 x 1023 N atoms
9. How many moles of Fe are there in a 275 gram sample?
275g Fe 1mol Fe
55.85g Fe
4.92mol Fe
10. Calculate the molecular mass/molar mass for each of the following?
a. Al(NO3)3
Al : 26.98 g
N: 14.01g
O 16.00g
NO3 : 14.01 + 3(16.00) = 62.01g
Al(NO3)3 : 26.98 g + 3(62.01g) = 213.01g/mol
b. NaCl
Na : 22.99g
Cl : 35.45g
NaCl : 58.44g/mol
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