In 3.00 x 10^20 molecules of c12h22o11, how many c atoms are presented?

3.6 x 10²¹ carbon atoms

Explanation:

Data Given:

C₁₂H₂₂O₁₁ = 3.00 x 10²⁰ molecules

Carbon atoms = ?

Solution:

Step 1.

First find number of moles of C₁₂H₂₂O₁₁

Formula used

             no. of moles = no. of molecules/ Avogadro's number

Put vales in above formula

             no. of moles =  3.00 x 10²⁰ / 6.23 x 10²³

             no. of moles =  4.82 x 10⁻⁴ mol

Step 2.

Now find mass of 4.82 x 10⁻⁴ moles of C₁₂H₂₂O₁₁  

Molar mass C₁₂H₂₂O₁₁ = 12(12) + 22(1) + 11(16)

Molar mass C₁₂H₂₂O₁₁ = 342 g/mol

Formula used

                no. of moles = mass in grams / Molar mass

Put values in formula

               4.82 x 10⁻⁴ mol =  mass in grams / 342 g/mol

Rearrange the above equation

              mass in grams =   4.82 x 10⁻⁴ mol x 342 g/mol

              mass in grams =   0.165 g

So,

C₁₂H₂₂O₁₁ = 0.165 g

Step 3.

calculate the percent composition of Carbon (C) in C₁₂H₂₂O₁₁

Since the percentage of compound is 100

So,

Formula used:

Percent Composition of Carbon (C) = mass of carbon  / molar mass x 100

Put values in formula

Percent Composition of Carbon (C) = 144  / 342 x 100

Percent of Carbon (C) = 42 %

It means that for ever gram of C₁₂H₂₂O₁₁ there is 0.42 g of C is present.

So,

For the 0.165 g of C₁₂H₂₂O₁₁ the mass of C will be

mass of Carbon (C) = 0.42 x 0.165 g

mass of Carbon (C) = 0.0693 g

Step 5.

Now find the number of moles for 0.0693 g of Carbon (C)

Molar mass of C = 12 g/mol

Formula Used

            no. of moles = mass in grams / Molar mass

put values in above formula

              no. of moles = 0.0693 g  / 12 g/mol

               no. of mol = 0.0058

no of moles of Carbon = 0.0058

Step 5.

Now find number of atoms in 0.0058 moles of carbon

Formula used

             no. of moles = no. of atoms of C / Avogadro's number

Put vales in above formula

             0.0058 = no. of atoms of C / 6.23 x 10²³

Rearrange the above equation

             no. of atoms of C =  0.0058 x 6.23 x 10²³

             no. of atoms of C =  3.6 x 10²¹

So,

3.6 x 10²¹ carbon atoms are present in 3.00 x 10²⁰ molecules of C₁₂H₂₂O₁₁

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