mols of H2O produced = 5.671 /18 = .315
mols of CO2 produced = 36.933 /44 =.8394
Cx Hy Oz + n O2 ----> .8394 CO2 + .315 H2O
well, y = .630 right off
and x = .8394
z + 2 n = 2(.8394) + .315 = 1.99
but
mass of O2 on left = 5.671+36.933-17.422
= 25.182 grams
which is 25.182/32 = .787 mols of O2 on left
so n = .787
z + 1.57 = 1.99
z = .416
so
I have
.8394 C
.630 H
.416 O
or
2.02 C
1.51 H
1 O
or
4 C
3 H
2 O or C2H3O2
which has mass/mol of 24+3+32 = 59
now 332.16/59 = 5.63
humm we have about 5 and a half mols so made a mistake somewhere. check my algebra !
An organic compound consists of only carbon, hydrogen, and oxygen. Through combustion analysis, 5.671g of water and 36.933g of carbon dioxide are formed from 17.422g of sample.
C:12.0 amu O: 16.0 amu H:1.0 amu
In the space provided please write the empirical formula and, if the compound has a molecular mass of 332.16 amu write the molecular formula?
3 answers
I'll show you how to do this the chemistry way.
Convert 5.671 g H2O to mols H atoms. That's 5.671 x (2 atomic mass H/molar mass H2O) = approx 0.630 but you need to go through all of these calculations and confirm that. Some I have estimated.
Convert 36.933 g CO2 to g C. That's
36.933 x (atomic mass C/molar mass CO2) = approx 10.07 g C
g O = g sample - g C - g O = 17.422 - 0.630 - 10.073 = approx 6.72
Now convert these g to mols.
mol C = 10.1/12 = 0.839
mols H = 0.630/1 = 0.63
mols O = 6.72/16 = 0.42
Find the ratio of these to each other with the smallest being no less than 1.00. The easy way to do that is to divide each number by the smallest number.
C = 0.839/0.42 = 1.997 which rounds to 2.0
H = 0.63/0.42 = 1.5
o = 0.42/0.42 = 1.00
So the ratio is 2:1.5:1. We want small WHOLE numbers so that is 4:3:2 and the empirical formula is C4H3O2 so Damon was right to this point. He made a typo on the next step and coped this as C2H3O2. Going with C4H3O2 the empirical mass is 48+3+32 = 83. The molar mass in the problem is 332; therefore,
empirical mass x ?# = molar mass
83 x #? = 332
#? = 332/83 = 4.0 so the molecular formula is
(C4H3O2)4 or C16H12O8
Convert 5.671 g H2O to mols H atoms. That's 5.671 x (2 atomic mass H/molar mass H2O) = approx 0.630 but you need to go through all of these calculations and confirm that. Some I have estimated.
Convert 36.933 g CO2 to g C. That's
36.933 x (atomic mass C/molar mass CO2) = approx 10.07 g C
g O = g sample - g C - g O = 17.422 - 0.630 - 10.073 = approx 6.72
Now convert these g to mols.
mol C = 10.1/12 = 0.839
mols H = 0.630/1 = 0.63
mols O = 6.72/16 = 0.42
Find the ratio of these to each other with the smallest being no less than 1.00. The easy way to do that is to divide each number by the smallest number.
C = 0.839/0.42 = 1.997 which rounds to 2.0
H = 0.63/0.42 = 1.5
o = 0.42/0.42 = 1.00
So the ratio is 2:1.5:1. We want small WHOLE numbers so that is 4:3:2 and the empirical formula is C4H3O2 so Damon was right to this point. He made a typo on the next step and coped this as C2H3O2. Going with C4H3O2 the empirical mass is 48+3+32 = 83. The molar mass in the problem is 332; therefore,
empirical mass x ?# = molar mass
83 x #? = 332
#? = 332/83 = 4.0 so the molecular formula is
(C4H3O2)4 or C16H12O8
Ive been working on a homework assignment for hours now. It's only 6 problems and I cant figure them out. I've tried and tried but everytime I type something in It says im wrong! Please help. A compound with a formula mass of 116.07 amu is found to be 41.39% carbon, 3.47% hydrogen and 55.14% oxygen by mass. Find its molecular formula
1 answer