Question

How many C₄H₁₀ molecules are contained in 9.213 g of this compound? How many hydrogen atoms?

Short answer

No of molecules of C₄H₁₀ will be $9.452\times {10}^{22}$

The no of atoms of hydrogen will be $9.452\times {10}^{23}$

Step-by-step solution

Determine molecular mass

To calculate no of molecules in 9.213 g of that compound, we have to evaluate the molar mass of that compound.

Element	Quantity		Molar mass (g/mol element)		Subtotal (g/mol compound)
C	4	×	12.01	=	48.04
H	10	×	1.008	=	10.08
			Molecular mass(g/mol)	=	58.12

Determine number of C4H10 molecules 

No. of molecules of that compound in 9.213 g of that compound,

$\begin{array}{r}9.213\mathrm{g}\mathrm{o}\mathrm{f}{\mathrm{C}}_4{\mathrm{H}}_{10}\times \frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{f}{\mathrm{C}}_4{\mathrm{H}}_{10}}{58.12\mathrm{g}\mathrm{o}\mathrm{f}{\mathrm{C}}_4{\mathrm{H}}_{10}}\times \frac{6.022\times {10}^{23}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}{\mathrm{C}}_4{\mathrm{H}}_{10}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{f}{\mathrm{C}}_4{\mathrm{H}}_{10}}\\ =9.452\times {10}^{22}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}{\mathrm{C}}_4{\mathrm{H}}_{10}\end{array}$

Hence, no of molecules of C₄H₁₀ will be $9.452\times {10}^{22}$

Determine number of hydrogen atom

No. of hydrogen atoms,

$9.452\times {10}^{22}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}{\mathrm{C}}_4{\mathrm{H}}_{10}\times \frac{10\mathrm{H}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}}{1\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}{\mathrm{C}}_4{\mathrm{H}}_{10}}=9.452\times {10}^{23}\mathrm{H}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}$

Hence, the no of atoms of hydrogen will be $9.452\times {10}^{23}$