Quiz 3 (08-09)
Chem 171
Name_____________________
1)
What is the molar mass, in grams, of Ca(HCO3)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 Al2(SO4)3 are in 210.0 g of that substance.
(2 mol Al)(27.00 g/mol) + (3 mol S)(32.06 g/mol) + (12 mol O)(16.00 g/mol) = 342.1 g/mol
210.0 g x mol/342.1g = 0.614 mol
3)
How many formula units of Al2(SO4)3 are in 210.0 of that substance?
path is mass to mol to formula units
23
23
210.0 g x mol/342.1g x 6.122 x 10 formula units/mol = 3.70 x 10
4)
What is the mass percent of Al in Al2(SO4)3?
(mass Al in 1 mol compound/mass 1 mol compound) x 100
2 mol A x 27.00 g/mol = 54.00 g
54.00 g/342.1g x 100 = 15.8 %
5)
What is the mass of S, in grams, in 210.0 g Al2(SO4)3?
path is mass compound to mol compound to mol S to mass S
210.g x mol/342.1g x 3 mol S/mol compound x 32.00 g/mol S = 58.94 g
6)
The empirical formula of apigenin, a yellow dye for wool is C3H2O. 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 C15H10O5 is the molecular formula
7)
A 0.0989 g sample of an alcohol (contains C, H, O) is burned in oxygen to yield 0.2160 g CO2 and 0.1194 g H2O.
What is the empirical formula of this acid?
Remember that all C ends up in CO2 and all H ends up in H2O
Path is mass CO2 to mole CO2 to mole C to mass C
Path is mass H2O to mole H2O to mole H to mass H
Mass of O is found by difference.
0.2160 g x mol CO2/44.0 g x 1 mol C/1 mol CO2 = 0.0049 mol C and mass is x 12.00 g/mol C = 0.0589 g C
0.1194 g x mol H2O/18.00 g x 2 mol H/1 mol H2O = 0.01326 mol H and mass is x 1.008 g/mol H = 0.01337 g H
Total mass = mass O + mass C + mass H so 0.1204 = mass O + 0.589 + 0.01337 and mass O is 0.07227 g
Need moles: so far know 0.0049 moles C and 0.01326 moles H, use 0.07227 g x mol/16.00 g = 0.00166 mol O
Divide by smallest number of moles (0.00166) to yield C3H8O
8)
9)
Balance the following equations:
a)
TiCl4
b)
Sn
+
+
2 H2O
→
TiO2
2 NaOH
→
Na2SnO2
+
4 HCl
+
H2
Hydrofluoric acid reacts with sodium silicate as indicated below:
Molar mass:
122.07
20.08
144.1
41.9
18.02
Na2SiO3 (s) + 8HF (aq) → H2SiF6(aq) + 2NaF (aq) + 3H2O (l)
a) How many grams of Na2SiO3 are consumed in a reaction with 35.0 g HF?
path is mass HF to mol HF to mol Na2SiO3 to mass Na2SiO3
35.0 g HF x mol HF/20.08 g x 1 mol Na2SiO3/8 mol HF x 122.07 g/mol Na2SiO3 = 26.6 g
b) If the reaction forms 87.6 g H2SiF6(aq), how many g NaF would be formed?
path is mass H2SiF6 to mol H2SiF6 to mol NaF to mass NaF
87.6 g x mol H2SiF6/144.1 g x 2 mol NaF/1 mol H2SiF6 x 41.9 g/mol = 50.9 g
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) + 2AgNO3(aq) → 2Ag (s) + Zn(NO3)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 AgNO3/1 mol Zh
b) Circle the limiting reactant? AgNO3
c) How much Ag(s) (in grams) could be formed?
Path is mass AgNO3 to mol AgNO3 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
Final Amount
Zn
0.0306
AgNO3
0.0147
Ag
0
- 0.00736
- 0.0147
+ 0.0147
0.0232
0
0.0147
Zn(NO3)
0