Question

An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 5.00g of this compound produced 7.33g of carbon dioxide and 3.00g of water. Question: How many moles of carbon, C, were in the original sample? Express your answer to three significant figures and include the appropriate units.

Answer 1

A combustion reaction involves an organic compound reacted with oxygen. The general chemical equation is as follows:

Organic Compound + Oxygen = CO2 + H2O

To calculate the amount of C present in the original sample, we use the values given and assume that there is complete combustion that is happening.

7.33 g CO2 ( 1 mol CO2 / 44.01 g CO2)(1 mol C / 1 mol CO2) = 0.167 mol C

Therefore, 0.167 mol of C was originally in the sample.

Answer 2

The number of moles of Carbon in 5.00g of a the unknown compound containing, carbon, hydrogen and oxygen (CxHyOz) is  0.167 moles.

Further Explanation

Hydrocarbons

• Hydrocarbons are organic compounds that are made up of hydrogen and carbon elements. However, some hydrocarbons such as alcohols and alkanoic acids contain oxygen element.

Combustion of hydrocarbons

• When hydrocarbons are burned in presence of oxygen gas they form carbon dioxide and water as the only products.

That is;

CxHy + O2 = CO2 + H2O

In this case;

• The Unknown compound contains carbon, hydrogen and oxygen, (CxHyOz)

• Therefore on combustion;

CxHyOz + O2 = CO2 + H2O  

• Mass of Carbon (iv) oxide produced = 7.33 g  
• Mass of water produced = 3.00 g  

Required to determine the number of moles of Carbon in the Unknown compound

Step 1: Mass of carbon in 7.33 g of Carbon (iv) oxide

1 mole of CO2 = 44 g  

1 Mole of CO2 = 12 g of carbon  

Therefore;

44 g of CO2 contain 12 g of carbon

Hence; mass of carbon in 7.33 g CO2  

= (7.33 x 12 )/ 44

= 1.99909 g

 

Step 2: Moles of Carbon  

Moles = Mass /RAM  

RAM of carbon = 12  

Number of moles = 1.99909 g/12 g/mol

                               = 0.16659  

                               = 0.167 Moles (3 sf)

Keywords: Combustion, moles, hydrocarbon, carbon (iv) oxide  

Learn more about:

• Empirical formula: brainly.com/question/11907106
• Example of empirical formula: brainly.com/question/12061700
• Molecular formula: brainly.com/question/12061700
• Hydrocarbons, molecular formula and empirical formula: brainly.com/question/11601506

Level: High school  

Subject: Chemistry  

Topic: Empirical and molecular formula