An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 6.50 g of this compound produced 9.53 g of carbon dioxide and 3.90 g of water. If 6.50 g of the unknown compound contained 0.217 mol of C and 0.433 mol of H, how many moles of oxygen, O, were in the sample?

There were 0.216 moles of oxygen in the sample

Explanation:

Step 1: Data given

Mass of compound = 6.50 grams

Mass of CO2 = 9.53 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 3.90 grams

Molar mass H2O = 18.02 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 9.53 grams / 44.01 g/mol

Moles CO2 = 0.217 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.217 moles CO2 we have 0.217 moles C

Step 4: Calculate mass C

Mass C = 0.217 moles C *12.01 g/mol

Mass C = 2.61 grams

Step 5: calculate moles H2O

Moles H2O = 3.90 grams / 18.02 g/mol

Moles H2O = 0.216 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.216 moles H2O we have 0.433 moles H

Step 7: Calculate mass H

Mass H = 0.433 moles * 1.01 g/mol

Mass H = 0.437 grams

Step 8: Calculate mass O

Mass O = 6.50 grams - 2.61 grams - 0.437 grams

Mass O = 3.453 grams

Step 9: Calculate moles O

Moles O = 3.453 grams / 16.0 g/mol

Moles O = 0.216 moles

There were 0.216 moles of oxygen in the sample

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

c. is the answer

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