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Question: A sample of \({{\text{C}}_{{\text{25}}}}{{\text{H}}_{{\text{52}}}}\) weights \(29.29\,{\text{g}}\). ...

A sample of C25H52{{\text{C}}_{{\text{25}}}}{{\text{H}}_{{\text{52}}}} weights 29.29 g29.29\,{\text{g}}. Its molar mass is 352.77 g/mol352.77\,{\text{g/mol}}. How many grams of carbon are in this sample?

Explanation

Solution

The mole concept is used to determine the amount of the substance which is obtained by taking the ratio of the mass of the substance to the molecular mass of the substance.
Here, we have to determine the grams of the carbon present in the one mole of the sample. After that, we can determine the mass of carbon in grams in a given amount of the sample.

Complete solution:
Here, the chemical formula of the compound is C25H52{{\text{C}}_{{\text{25}}}}{{\text{H}}_{{\text{52}}}}.and the molar mass of the compound is352.77 g/mol352.77\,{\text{g/mol}}.
Here, the chemical formula indicates one mole of the C25H52{{\text{C}}_{{\text{25}}}}{{\text{H}}_{{\text{52}}}} has 25 carbons atoms.
The atomic mass of carbon is 12.011 amu{\text{12}}{\text{.011}}\,{\text{amu}} that is mass is in atomic mass unit hence, the weight of the carbon in a sample of C25H52{{\text{C}}_{{\text{25}}}}{{\text{H}}_{{\text{52}}}} in grams is calculated as follows:
12.011×25 = 300.275 g{\text{12}}{\text{.011}} \times {\text{25 = 300}}{\text{.275}}\,{\text{g}}
Thus, the mass of carbon in a mole of the sample is 300.275 g{\text{300}}{\text{.275}}\,{\text{g}}.
One mole of the sample has 352.77 g352.77\,{\text{g}} mass.
Therefore, the mass of carbon in 352.77 g352.77\,{\text{g}} of sample is 300.275 g{\text{300}}{\text{.275}}\,{\text{g}}.
Now, we have to determine the mass of carbon in 1 gram of the sample as follows:
352.77 g of sample  =  300.275 g of C{\text{352}}{\text{.77}}\,{\text{g}}\,{\text{of}}\,{\text{sample}}\,{\text{ = }}\,{\text{300}}{\text{.275}}\,{\text{g}}\,{\text{of}}\,{\text{C}}
1 g of sample  =  300.275 g of C352.77 g of sample×1 g of sample{\text{1}}\,{\text{g}}\,{\text{of}}\,{\text{sample}}\,{\text{ = }}\,\dfrac{{{\text{300}}{\text{.275}}\,{\text{g}}\,{\text{of}}\,{\text{C}}}}{{{\text{352}}{\text{.77}}\,{\text{g}}\,{\text{of}}\,{\text{sample}}}} \times {\text{1}}\,{\text{g}}\,{\text{of}}\,{\text{sample}}
1 g of sample  =  0.851192 g of C{\text{1}}\,{\text{g}}\,{\text{of}}\,{\text{sample}}\,{\text{ = }}\,0.851192\,{\text{g}}\,{\text{of}}\,{\text{C}}
Now, we have to determine the mass of the carbon in 29.29 g29.29\,{\text{g}} as follows:
1 g of sample  =  0.851192 g of C{\text{1}}\,{\text{g}}\,{\text{of}}\,{\text{sample}}\,{\text{ = }}\,0.851192\,{\text{g}}\,{\text{of}}\,{\text{C}}
29.29 g of sample  =  0.851192 g of C1 g of sample×29.29 g of sample29.29\,{\text{g}}\,{\text{of}}\,{\text{sample}}\,{\text{ = }}\,\dfrac{{0.851192\,{\text{g}}\,{\text{of}}\,{\text{C}}}}{{{\text{1}}\,{\text{g}}\,{\text{of}}\,{\text{sample}}}} \times 29.29\,{\text{g}}\,{\text{of}}\,{\text{sample}}
29.29 g of sample  =  24.9314 g of C29.29\,{\text{g}}\,{\text{of}}\,{\text{sample}}\,{\text{ = }}\,24.9314\,{\text{g}}\,{\text{of}}\,{\text{C}}
Thus, the mass of the carbon present in the sample in grams is 24.93 g24.93\,{\text{g}}.

Note: Molecular weight is the average mass of the molecules and it is deferrable from molecule to molecule. In the case of the atom, it is called atomic mass while in the case of the molecular mass it is calculated by taking the sum of the atomic masses of all the atoms in the molecule.
The SI unit of the molecular weight is gram per mole. Its unit itself indicates the ratio of the mass of the substance to the molar molecular mass of the substance.