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?
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 !
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
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
To determine the empirical formula and molecular formula of the organic compound, we need to calculate the ratio of the elements present in the compound using the given information.
First, we need to determine the moles of water formed. The molar mass of water (H₂O) is 18.0 g/mol. We can calculate the moles of water by dividing the mass of water formed (5.671 g) by its molar mass:
Moles of water = 5.671 g / 18.0 g/mol = 0.3151 mol
Next, we need to determine the moles of carbon dioxide formed. The molar mass of carbon dioxide (CO₂) is 44.01 g/mol. We can calculate the moles of carbon dioxide by dividing the mass of carbon dioxide formed (36.933 g) by its molar mass:
Moles of carbon dioxide = 36.933 g / 44.01 g/mol = 0.8397 mol
Now, we need to calculate the moles of carbon and moles of hydrogen present in the given sample. The remaining mass of the sample after subtracting the mass of water and carbon dioxide is:
Mass of sample - (Mass of water + Mass of carbon dioxide) = 17.422 g - (5.671 g + 36.933 g) = -25.182 g
Since we cannot have a negative mass, this suggests that there was an error in the measurements or calculations. Please check the values and ensure they are correct.
If we assume that the mass of the sample should be positive, we can proceed with the calculations using the given values, but note that the results would not be accurate.
We will continue the calculations using the given values, but please double-check the numbers provided.
Next, we need to calculate the moles of carbon and hydrogen in the compound.
Moles of carbon = -25.182 g / 12.0 g/mol = -2.0985 mol
Moles of hydrogen = -25.182 g / 1.0 g/mol = -25.182 mol
As mentioned earlier, these negative values are not valid, so please verify the values provided.
Once we have determined the correct values for the moles of carbon and hydrogen, we can calculate the empirical formula. To do this, we divide each element's moles by the smallest number of moles. However, since we don't have valid mole values, we cannot calculate the empirical formula accurately.
Without the accurate empirical formula, we cannot calculate the molecular formula using the molecular mass mentioned (332.16 amu) as it requires the empirical formula for comparison.
In summary, please double-check the given values for the masses of water, carbon dioxide, and the sample. Once you have accurate values, divide them by molar masses to calculate the moles of each compound. From there, you can determine the empirical formula by dividing the moles of each element by the smallest number of moles. This will allow you to calculate the molecular formula using the molecular mass.