CH 115 Fall 2014Worksheet 19 Phenyl magnesium bromide is used

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CH 115 Fall 2014
Worksheet 19
1. Phenyl magnesium bromide is used as a Grignard reagent in organic synthesis.
Determine its empirical and molecular formula if its molar mass is 181.313 g/mol
and it contains 39.7458% C, 2.77956% H, 13.4050% Mg, and a unknown percentage
of Br.
All percentages must add up to 100 – can find the missing percentage of Br.
100-55.93036 = 44.06964% Br
Assume 100 g of sample  39.7458 g C, 2.77956 g H, 13.4050 g Mg, 44.06964 g
Br
Convert grams to moles.
39.7458 g C / 12 g/mol = 3.309392173 mol C
2.77956 g H / 1 g/mol = 2.77956 mol H
13.4050 g Mg / 24.31 g/mol = 0.5514191691 mol Mg
44.06964 g Br / 79.9 g/mol = 0.5515599499 mol Br
Divide through by the lowest number of moles.
C=6
H=5
Mg = 1
Br = 1
Empirical formula = C6H5MgBr
To find the molecular formula, first calculate the molar mass of the empirical
formula.  181.27 g/mol
Divide the molar mass of the molecular formula by the empirical formula to
calculate the factor you must multiply the empirical formula by. In this case,
181.313/181.27 = 1, so the molecular formula = C6H5MgBr
2. A 3.585-gram sample of a hydrocarbon contains 1.388 grams of C, 0.345 grams of H,
and 1.850 grams of O. Its molar mass is 62 g/mol. Determine its empirical and
molecular formula.
No percentages given – first need to calculate the percentages of C, H, and O
present in this sample.
1.388 g C/3.585 g = 38.71687587% C
0.345 g H/3.585 g = 9.623430962% H
1.850 g O/3.585 g = 51.60390516% O
Now, assume 100 g and follow the rest of the steps.
38.71687587 g C / 12 g/mol = 3.226406323 mol C
9.623430962 g H / 1 g/mol = 9.623430962 mol H
51.60390516 g O / 16 g/mol = 3.225244073 mol O
Divide through by least number.
C=1
H=3
O=1
Empirical formula = CH3O
Molar mass of empirical formula = 31 g/mol
62/31 = 2  molecular formula = C2H6O2
CH 115 Fall 2014
3. Balance the following chemical equations.
a). Br2 (l) + 2 H2O (l) + SO2 (g)  2HBr (aq) + H2SO4 (aq)
b). 2MnO2 + 2 H2SO4  2MnSO4 + O2 + 2H2O
c). 2FeCl3 + Be3(PO4)2  3BeCl2 + 2FePO4
d). 2C2H3O2Br + 3O2  4CO2 + 2H2O + 2HBr
Worksheet 19
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