Question

When a 7.40-mg sample of a compound containing carbon is burned completely, 18.6 mg of carbon dioxide is produced. What is the mass percentage of carbon in the compound?

Answer 1

Answer:

68.6 % of the compound is carbon

Explanation:

Step 1: Data given

Mass of 7.40 mg = 0.0074 grams

Mass of carbon dioxide = 18.6 mg = 0.0186 grams

Molar mass CO2 = 44.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 0.0186 grams / 44.01 g/mol

Moles CO2 = 4.23*10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 4.23*10^-4 moles CO2 we have 4.23*10^-4 moles C

Step 4: Calculate mass C

Mass C = 4.23*10^-4 moles * 12.01 g/mol

Mass C = 0.00508 grams

Step 5: calculate the mass percentage of carbon

% C= (0.00508 grams / 0.0074 grams ) * 100 %

% C = 68.6 %

68.6 % of the compound is carbon

Answer 2

Answer: The mass percent of carbon in sample is 68.51 %

Explanation:

We are given:

Mass of $CO_2=18.6mg=0.0186g$

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.0186 g of carbon dioxide, $\frac{12}{44}\times 0.0186=0.00507=5.07mg$ of carbon will be contained.

To calculate the mass percentage of carbon in sample, we use the equation:

$\text{Mass percent of carbon}=\frac{\text{Mass of carbon}}{\text{Mass of sample}}\times 100$

Mass of carbon = 5.07 mg

Mass of sample = 7.40 mg

Putting values in above equation, we get:

$\text{Mass percent of carbon}=\frac{5.07mg}{7.40mg}\times 100=68.51\%$

Hence, the mass percent of carbon in sample is 68.51 %