Calculate the number of C. H. and O atoms in 1.50 g of

Step 1 of 3

Here, we are going to calculate the number of C. H. and O atoms in 1.50 g of glucose. \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\).

We know that,

Molar mass of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)=(6 \times 12.01 \mathrm{~g})+12 \times 1.008 \mathrm{~g}+6 \times 16.00 \mathrm{~g}=180.2 \mathrm{~g} / \mathrm{mol}\)

1.0 mol of glucose contains \(=6.022 \times 10^{23}\) molecules of glucose

The conversion factors are,

\(\frac{6.022 \times 10^{23} \text { molecules of glucose }}{1.0 \text { molof glucose }} \text { and } \frac{1.0 \text { mol of glucose }}{180.2 \text { gof glucose }} \text { and } \frac{6.0 \text { Catoms of glucose }}{1 \text { molecule of glucose }}\)

The number of C in glucose

\(1.50 \mathrm{~g} \times \frac{6.022 \times 10^{23} \text { molecules of glucose }}{1.0 \text { molof glucose }} \times \frac{1.0 \text { mol of glucose }}{180.2 \text { gof glucose }} \times \frac{6.0 \text { Catoms of glucose }}{1 \text { molecule of glucose }}\)

\(=3.01 \times 10^{22} \text { atoms of } \mathrm{C} \text { in glucose }\)

Thus, we have found that 1.50 g of glucose contains \(3.01 \times 10^{22}\) atoms of Carbon.