Question

ioxide and 4.50 of water.
1) How many moles of carbon, C, were in the original sample?
An unknown compound contains only carbon, hydrogen, and oxygen (). Combustion of 7.50 of this compound produced 11.0 of carbon dioxide and 4.50 of water.

1) How many moles of carbon, C, were in the original sample?

Answer 1

1) How many moles of carbon, C, were in the original sample?

We are given the amount carbon dioxide being produced from the reaction. We assume that the reaction is complete and the only product that contain carbon is only carbon dioxide. We calculate as follows:

11.0 g CO2 ( 1 mol / 44.01 g) ( 1 mol C / 1 mol CO2) = 0.30 mol C

Answer 2

7.50 g of a compound that contains hydrogen, carbon and oxygen, which produces 11.0 g of carbon dioxide and 4.5 g of water when burned, contains 0.25 moles of carbon in the original sample.

Further Explanation:

A compound  

• A compound is a substances that contains two or more different atoms that are bonded together.
• When the atoms are similar the substance is known as a molecule, therefore not all molecules are compounds.

Hydrocarbons  

• Hydrocarbons are compounds that contains hydrogen and carbon as the only elements. Other hydrocarbons are composed of hydrogen, carbon and oxygen elements.
• When hydrocarbons are burned in air, they form carbon dioxide and water:

That is;

CxHyOn + O2  = CO2 + H2O  

Moles of carbon in the original sample of a Hydrocarbon  

• We can calculate the original number of moles in the original sample of the hydrocarbon.

In our case:

Mass of CO2 formed is 11.0 g  

Mass of the hydrocarbon is 7.5 g  

Mass of carbon:  

1 mole of CO2 contains 12.0 g of carbon

44.0 g/mol contains 12.0 g of carbon

Hence; 11.0 g of CO2 will contain;

= (11x 12)/44.0g

= 3 g of carbon  

This means; the original mass of Carbon in 7.5 g of the hydrocarbon is 3 g  

Number of moles = mass/R.A.M

                             = 3 g/ 12.0 g/mol

                             = 0.25 moles  

Hence; 7.5 g of the hydrocarbon contained 0.25 moles of carbon.

Learn more about:

• A compound: brainly.com/question/1841136
• Molecular formula: brainly.com/question/5366502
• Empirical Formula: brainly.com/question/5366502
• Example Question: brainly.com/question/11044214

Level: High school  

Subject: Chemistry  

Topic: Empirical formula and molecular formula