Q23P - Chemistry: Molecular Nature Of Matter And Change

Question

Cisplatin(right) or platinol, is used in the treatment of certain cancers. 

Calculate

1. The moles of compound in 285.3 g of cisplatin;
2. The number of hydrogen atoms in 0.98 mol of cisplatin.

Step-by-Step Solution

Answer

1. The moles of compound in 285.3 g of cisplatin is 0.951 mol
2. The number of hydrogen atoms in 0.98 mol of cisplatin is $\mathbf{3}\mathbf{.}\mathbf{54}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{atoms}$

Step 1: Definition of moles

A standard scientific unit for measuring large quantities of very small entities, such as atoms, molecules or other specified particles

Step 2: Calculations of moles of compound in 285.3 g of cisplatin

The molar mass of cisplatin is 300.06 $\frac{\mathbf{g}}{\mathbf{mol}}$

Moles of cisplatin $\mathbf{=}\mathbf{285}\mathbf{.}\mathbf{3}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{cisplatin}\left(\frac{\mathrm{molc} \mathrm{is} \mathrm{platin}}{300.06 g c \mathrm{is} \mathrm{platin} }\right)$

Moles of cisplatin $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{951}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{c}\mathbf{ }\mathbf{isplatin}$

Step 3: Calculations of number of hydrogen atoms in 0.98 mol of cisplatin

Number of H atoms $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{98}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{cisplatin}\left(\frac{6\mathrm{molH} \mathrm{atoms}}{1\mathrm{mol} \mathrm{cisplatin}}\right)\left(\frac{6.022\times {10}^{23}}{\mathrm{molH} \mathrm{atoms}}\right)$

Number of H atoms $\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{54}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{ }\mathbf{atoms}$