Chem Help~~!!!!!????

Question

A 1.20 g sample of a compound gave 2.92 g of CO2 and 1.22 g of H2O on combustion in oxygen. The compound is known to contain only C, H, and O. What is its simplest formula?

Answer 1

The carbon and hydrogen in the combustion products can only have come from the original compound.  Once we account for that, any left over material in the original is oxygen.  

In CO2, 12 of 44 grams is carbon, so g of carbon=

12x 2.92/44

In H2O, 2 of 18 grams is water, so g of hydrogen=

2 x 1.22/18

Once you have the grams of each element in the compound combusted, you divide each amount by its atomic weight.  This will give you the ratio of moles of C,H and O, from which you can derive the simplest formula.  

Answer 2

To work this out you need to know the numbe of moles of each element. All of the C and H that were in the original mass are now in the products.

Mass of CO2 = 2.92 g

moles CO2 = mass / molecular mass

molecular mass CO2 = 12.01 + (2 x 16.0) = 44.01 g/mol

therefore moles CO2 = 2.92 g / 44.01 g/mol

= 0.0663 moles

In 0.0663 moles of CO2 there are 0.0663 moles of C

Mass of C in product = moles x molecular weight

= 0.0663 x 12.01

= 0.7962 g of Carbon in product

Mass H2O = 1.22 g

Molecular weight H2O = (1.008 x 2) + 16.00 = 18.016 g/mol

therefore moles H2O = 1.22 g / 18.016 g/mol

= 0.0677 moles H20

In 0.0677 moles of H2O there are (2 x 0.677 moles) of H

= 0.1354 moles of H in the products

mass H in product = moles x molecular weight

= 0.1354 * 1.008

= 0.1364 g of H in products (H2O)

So, all the C and H from the product were from the initial sample. To calculate the mass of O from the initial sample is

mass original sample - mass C - mass H

= 1.20 g - 0.7962 g - 0.1364 g

= 0.2674 g of O were in the original sample.

moles of O in original sample = mass/ molecular weight

= 0.2674 g / 16.00 g/mol

= 0.01671 moles of O

(The rest of the O came from the O2 in the combustion reaction)

Now, to work out empirical formula you compare the number of moles of C : H : O

0.0663 : 0.0677 : 0.01671

Divide by the lowest amount to get the lowest whole ratio

C:H:O

(0.0663 / 0.01671) : (0.0677 / 0.01671) : (0.01671/0.01671)

3.97 : 4.051 : 1

Since C and H are around 4 we can say that the empirical formula is

C4H4O