ANSWERS to examples worked in class
1. Chemical Stoichiometry - per cent composition (uses mass information)
A. 4.421 g cmpd (Ca and Cl)
-1.593 g Ca per cent is Parts per Hundred
--------- % X = 100*(mass X)/(total mass)
2.828 g Cl
[1.593]
% Ca = ------ x 100 = 36.04 b/c there are only two components
[4.421] present in this compound, % Cl
can be obtained from 100 - 36.04,
or by 100*(2.828)/(4.421).
B. 21.38 g oxide (Hg and O)
-19.67 g Hg [19.67]
--------- % Hg = ------ x 100 = 92.00
1.71 g oxygen [21.38]
because this is a binary compound, % O = 100 - 92.00 = 8.00
or, % O = 100*(1.71)/(21.38)
C. i. burning 0.539 g sample forms 1.361 g carbon dioxide
All C in 1.361 g CO2 came from the sample.
How much C is present in 1.361 g CO2 ?
CO2 has a mass of 44 amu, due to 12 amu C, and 32 amu O.
1.361 g CO2[12 amu C]
? g C = -------------------- = 0.372 g C
[44 amu CO2]
ii. burning 0.539 g sample forms 0.239 g water.
All H in 0.239 g H2O came from the sample.
How much H is present in 0.239 g H2O ?
H2O has a mass of 18 amu, due to 2 amu H, and 16 amu O.
0.239 g H2O[2 amu H]
? g C = ------------------- = 0.026 g H
[18 amu H2O]
iii. what mass of sample is due to oxygen?
0.539 g cmpd ( C, H, and O )
-0.398 g C and H (from 0.372 g C + 0.026 g H, found above)
----------
0.141 g O
iv. find mass per cent composition [100x(part/whole)]
% C = 100*(0.372)/(0.539) = 69.02
(Oc course, sum of
% H = 100*(0.026)/(0.539) = 4.82 all percentages
must add to 100.)
% O = 100*(0.141)/(0.539) = 26.16
2. Chemical Stoichiometry - empirical formulas (shows counting information)
Empirical formulas show the simplest ratio of number of moles.
A. i. convert from masses to moles
Assume all percentages are mass percentages, unless otherwise stated.
So 91.30 % C is same as saying 91.30 parts carbon per 100 parts compound,
which is the same as saying 91.30 grams carbon per 100 grams compound.
This latter reading allows for ready conversion from mass to mole...
91.30 g C[1.0 mole C]
? mole C = -------------------- = 7.608 mole C
[12.0 g C]
% H = 100 - 91.30 % C = 8.70 % H, same as saying 8.70 g hydrogen
per 100 grams of compound.
Convert to mole hydrogen...
8.70 g H[1.0 mole H]
? mole H = ------------------- = 8.70 mole H
[1.0 g H]
ii. calculate simplest ratio of moles. Divide all
calculated mole values by lowest value. Lowest mole value
is for mole C. So, divide mole C, and mole H, by mole C...
mole H 8.70 mole C 7.608
------- = ----- = 1.1435.. ------- = ----- = 1.00...
mole C 7.608 mole C 7.608
Using this information, the formula could be represented as
C1.00...H1.14435...
However, proceed to find integer subscripts by multiplying
with unit fractions (as was done in Law of Multiple Proportions)..
multiply by mole C mole H
1/1 1 1.14435...
2/2 2 2.28870...
...
7/7 7 8.0047...
and report formula as: C7H8
B. i. Given, % Sr = 48.78, % N = 15.59, find per cent oxygen...
100.00 (grams or parts of compound, Sr + N + O)
-64.37 (grams or parts of Sr + N)
----------
35.63 (grams or parts of oxygen)
ii. convert gram masses to moles...
48.78 g Sr[1 mole Sr]
? mole Sr = -------------------- = 0.5567 mole Sr
[87.62 g Sr]
15.59 g N [1 mole N]
? mole N = ------------------- = 1.1136 mole N
[14 g N]
35.63 g O[1 mole O]
? mole O = ------------------ = 2.2269 mole O
[16 g O]
iii. divide all mole values, by lowest mole value, i.e.,
divide all by 0.5567
rel.mole Sr = (0.5567 / 0.5567) = 1 Sr
rel.mole N = (1.1136 / 0.5567) = 2.0002... N
rel.mole O = (2.2269 / 0.5567) = 3.9999... O
iv. report empirical formula as: SrN2O4
C. i. Given compound is composed of elements, C, H, and O, and has
a molecular weight of 172 g/mol.
ii. 2.135 g of cmpd burns to form 5.462 g of carbon dioxide.
All carbon in the 5.462 g CO2 comes from the original cmpd.
So, how much carbon is present in 5.462 g CO2?
(CO2 has a MW = 44 amu, due to 12 amu C, and 32 amu O.)
5.462 g CO2[12 g C]
? g C = ------------------ = 1.4896 g C
[44 g CO2]
iii. 2.135 g of cmpd burns to form 2.234 g water.
All hydrogen in the 2.234 g water comes from the original cmpd.
So, how much hydrogen is present in 2.234 g H2O?
(H2O has a MW = 18 amu, due to 2 amu H, and 16 amu O.)
2.234 g H2O[2 g H]
? g C = ------------------ = 0.2482 g H
[18 g H2O]
iv. calculate mass oxygen in the cmpd, by difference...
2.135 g cmpd (C + H + O)
-1.739 g (C + H, from above)
----------
0.397 g oxygen
v. convert to mole
1.4896 g C[ 1 mole C]
? mole C = -------------------- = 0.1241 mole C
[12 g C]
0.2482 g H[ 1 mole H]
? mole H = -------------------- = 0.2482 mole H
[1 g H]
0.397 g O[ 1 mole O]
? mole C = -------------------- = 0.02481 mole O
[16 g O]
vi. divide all moles by lowest mole value, i.e., divide all
moles by 0.02481...
rel.mole C = (0.1241 / 0.02481) = 5.0029.. C
rel.mole H = (0.2482 / 0.02481) = 10.0039..H
rel.mole O = (0.02481/0.02481) = 1 O
vii. report empirical formula as: C5H10O
viii. calculate (true) molecular formula, given mol.wt. = 172 g/mol
Molecular formula is a whole number multiple of the empirical formula.
To find this multiple, compare mass of atoms in empirical formula,
with mass of molecular formula (i.e., 172)...
calculate mass of atoms in empirical formula ...C5H10O
5(12) + 10(1) + 16 = 86 amu, empirical formula wt.
compare to molecular formula wt. of 172 (i.e., set-up a ratio for example)
172
----- = 2 The empirical formula wt. goes into the molecular
86 wt twice. Conclusion: the molecular formula must
be composed of TWO empirical formulas.
So report molecular formula as C10H20O2
3. Chemical Stoichiometry - other applications for ratios of mass of
A. i. ? g carbon are contained in 57.83 g C8H18O2
MW = 8(12) + 18(1) + 2(16) = 146 amu, of which 96 amu is carbon
mass ratio of ( 96 C / 146 cmpd ) is the conversion factor sought here
57.83 g cmpd[96 g C]
? g C =------------------- = 44.06 g C
[126 g cmpd]
ii. ? g cmpd can be prepared from 14.275 g of hydrogen
MW = 8(12) + 18(1) + 2(16) = 146 amu, of which 18 amu is hydrogen
mass ratio of ( 18 H / 146 cmpd ) is the conversion factor sought here
14.275 g H[146 g cmpd]
? g cmpd = --------------------- = 115.79 g cmpd
[18 g H]
B. ? g S will combine with 4.829 g sodium to form NaHSO4
FW = 23(1) + 1(1) + 1(32) + 4(16) = 120 amu,
of which 23 amu is sodium, and 32 amu is sulfur.
mass ratio of (23 Na / 32 S ) is the conversion factor sought here
4.829 g Na[32 g S]
? g S = ----------------- = 6.719 g sulfur
[23 g Na]
C. ? g HgO will produce 2.594 g oxygen
Fw = 1(200.6) + 16 = 216.6 amu, of which 16 amu is oxygen.
the ratio of (16 O / 216.6 HgO) is the conversion factor sought here
2.594 g O[216.6 g HgO]
? g HgO = --------------------- = 35.116 g HgO
[16 g O]