Question

When 4.748 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 16.05 grams...

When 4.748 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 16.05 grams of CO2 and 3.286grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

empirical formula = ?
molecular formula = ?

Homework Answers

Answer #1

Empirical formula is the least coefficient formula, that is

CxHy

we must find x,y via gravimetry.

Typically we do this relating to moles of C,H

mol of CO2 = mass of CO2/MW of CO2 = (16.05)/(44) = 0.3647 mol of CO2

1 mol of CO2 = 1 mol of C ---> 0.3647 mol of CO2 = 0.3647 mol of C

mol of H2O = mass of H2O/MW of H2O= (3.286)/(18) = 0.1825 mol of H2O

1 mol of H2O= 2 mol of H ---> 0.1825 mol of H2O= 0.1825*2 = 0.365 mol of H

Ratios

H:C =0.365 / 0.3647 = 1:1

so

CH is the empirical formula

if MW 26.04

MW of empirical = 12 + 1 = 13

then

ratio = 2x

C2H2 is the sample

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
When 3.036 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 10.26 grams...
When 3.036 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 10.26 grams of CO2 and 2.101 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
When 2.050 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 6.280 grams...
When 2.050 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 6.280 grams of CO2 and 3.000grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 86.18 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. empirical formula =? molecular formula =?
When 3.793 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 12.34 grams...
When 3.793 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 12.34 grams of CO2 and 3.791grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
When 3.795 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 13.03 grams...
When 3.795 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 13.03 grams of CO2 and 2.134 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 128.2 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
In a combustion analysis of a hydrocarbon, 11.0 grams of CO2 and 6.78 grams of H2O...
In a combustion analysis of a hydrocarbon, 11.0 grams of CO2 and 6.78 grams of H2O are produced. What is the empirical formula of the hydrocarbon?
A 2.500-g sample of a hydrocarbon known to contain C, H, and O was burned in...
A 2.500-g sample of a hydrocarbon known to contain C, H, and O was burned in a C-H combustion train, producing 5.235 g of CO2 and 1.072 g of H2O. The molar mass of the compound is 126 g/mol. What is the molecular formula of the compound?
1.A 0.5538 g sample of a pure soluble bromide compound is dissolved in water, and all...
1.A 0.5538 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 1.1298 g. What is the mass percentage of bromine in the original compound? % 2. A student determines the manganese(II) content of a solution by first precipitating it as manganese(II) hydroxide, and then decomposing the hydroxide to...
5- Elemental analysis is sometimes carried out by combustion of the sample. For a hydrocarbon, the...
5- Elemental analysis is sometimes carried out by combustion of the sample. For a hydrocarbon, the only products formed are CO2 and H2O. If a 1.36-g sample of an unknown hydrocarbon is burned and 2.21 g of H2O are produced along with 4.07 g of CO2, what is the empirical formula of the hydrocarbon?
3.782 g of a substance (containing only C, H, and O) is burned in a combustion...
3.782 g of a substance (containing only C, H, and O) is burned in a combustion analysis apparatus. The mass of CO2 produced is 4.160 g, and the mass of H2O produced is 1.135 g. What is the empirical formula of the compound? In the empirical formula the coefficients must be reduced to the smallest possible integers. The convention for entering the empirical formula is to give C first followed by H then by any other elements in alphabetical order....
Combustion analysis of a compound yielded 269.01 g CO2, 55.09 g H2O, 70.30 g NO2, and...
Combustion analysis of a compound yielded 269.01 g CO2, 55.09 g H2O, 70.30 g NO2, and 97.90 g SO2. (a) What is the empirical formula of the compound? (Assume it contains no oxygen.) (b) If the molar mass of the compound is 196.30 g/mol, what is the molecular formula of the compound?