Key
1) What is the molar mass, in grams, of Ca(HCO
3
)
2
?
(1 mol Ca)(40.08 g/mol) + (2 mol H)(1.008 g/mol) + (2 mol C)((12.00 g/mol) + (6 mol O)(16.00 g/mol)
= 162.1 g/mol
2) How many moles of Ca(HCO
3
)
2
are in 156.0 g of that substance.
156 g x (mol/162.1 g) = 0.962 mol
3) How many formula units of Ca(HCO
3
)
2
are in 156.0 g of that substance? path: mass to moles to formula units
156 g x (mol/162.1 g) x 6.022 x 10
23
formula units/mol = 5.8 x 10
23
formulat units
4) What is the mass percent of C in Ca(HCO
3
)
2
?
2 mol C/mol compound so mass of C is 2 mol x 12.00 g/mol = 24 g mass C in 1 mol/mass 1 mol compound = 24.00 g/162.1 g x 100 = 14.8%
5) What is the mass of O, in grams, in 156.0 g Ca(HCO
3
)
2
? path is mass compound to moles compound to moles O to mass O
156 g x (mol/162.1 g) x 6 mol O/mol compound x 16.0 g/mol O = 92.4 g
6) The empirical formula of apigenin, a yellow dye for wool is C
3
H
2
O. The molecular mass is 270 g. What is the molecular formula of apigenin? molecular mass/empirical mass = n; use n to multiply subscripts empirical mass is (3 mol C)(12 g/mol) + (2 mol H)(1.008 g/mol) + (1 mol O)(16.00 g/mol) = 54.0 g
270g/mol divided by 54 g/mol = 5 so C
15
H
10
O
5
is the molecular formula
7) A 0.1204 g sample of an acid (contains C, H, O) is burned in oxygen to yield 0.2147 g CO
2
and 0.0884 g H
2
O.
What is the empirical formula of this acid?
Remember that all C ends up in CO
2
and all H ends up in H
2
O
Path is mass CO
2
to mole CO
2
to mole C to mass C
Path is mass H
2
O to mole H
2
O to mole H to mass H
Mass of O is found by difference.
0.2147 g x mol CO
2
/44.0 g x 1 mol C/1 mol CO
2
= 0.0049 mol C and mass is x 12.00 g/mol C = 0.0586 g C
0.0884 g x mol H
2
O/18.00 g x 2 mol H/1 mol H
2
O = 0.0098 mol H and mass is x 1.008 g/mol H = 0.00989 g H
Total mass = mass O + mass C + mass H so 0.1204 = mass O + 0.586 + 0.00989 and mass O is 0.0519 g
Now need moles: so far know 0.0049 moles C and 0.00989 moles H, use 0.0519 g x mol/16.00 g = 0.0032 mol O
Divide by smallest number of moles (0.0032) to yield C
1.5
H
3
O – multiply by 2 to get integers C
3
H
6
O
2
8) Balance the following equations: a) 4 Al + 3 O
2
→ 2 Al
2
O
3 b) Sn + 2 NaOH → Na
2
SnO
2
+ H
2
9) Lead (II) oxide reacts with ammonia as follows:
Molar mass: 223.2 17.02 207.2 28.0
3 PbO (s) + 2 NH
3
(g) → 3 Pb(s) + N
2
(g) + 3 H
18.02
2
O (l) a) How many grams of NH
3
are consumed in a reaction with 75.0 g PbO? path is mass PbO to mol PbO to mol NH
3
to mass NH
3
75.0 g x mol PbO/223.2g x 2 mol NH
3
/3 mol PbO x 17.02 g/mol NH
3
= 3.59 g b) If the reaction forms 56.4 g Pb(s), how many g N
2
would be formed? path is mass Pb to mol Pb to mol N
2
to mass N
2
56.4 g x mol Pb/207.2 g x 1 mol N
2
/3 mol Pb x 28.02 g/mol N
2

10) A strip of Zn metal, weighing 2.00 g, is placed in an aqueous solution containing 2.50 g of silver nitrate causing the following reaction to occur. Zn(s) + 2AgNO
3
(aq) → 2Ag (s) + Zn(NO
3
)
2
(aq) a) Complete the table below for this reaction. convert mass to moles for initial amounts use stoichiometric ratios to determine loss of each. Note ratio of 2 mol AgNO
3
/1 mol Zh b) Circle the limiting reactant? AgNO
3 c) How much Ag(s) (in grams) could be formed?
Path is mass AgNO
3
to mol AgNO
3
to mol Ag to mass Ag to get 0.0147 mol or 1.58 g d) How much of the excess reagent is left if the reaction goes to completion?
Table shows 0.0232 mol x 65.37 g/mol = 1.52 g e) What is the percent yield if 0.964 g of Ag(s) was formed.
Actual yield/Theoretical yield x 100
From actual calculation: 0.964/1.58 g x 100 = 61%
Using value given in class 0.0964/1.54 x 100 = 62.6 %
Initial Amount
AgNO
3
0.0147
Ag
0
Final Amount
Zn
0.0306
- 0.00736
0.0232
- 0.0147
0
+ 0.0147
0.0147
Zn(NO
0
3
)