Q24P - Chemistry: Molecular Nature Of Matter And Change

Question

Allyl sulfide gives garlic its characteristic odor.

Calculate 

1. The mass in grams of 2.63 mol of allyl sulfide;
2. The number of carbon atoms in 35.7g of allyl sulfide.

Step-by-Step Solution

Answer

1. The mass of allyl sulfide is $300.42 \mathrm{g} \mathrm{allyl} \mathrm{sulfide}$
2. The number of carbon atoms in 35.7g of allyl sulfide is $1.13\times {10}^{24}\mathrm{C} \mathrm{atoms}$

Step 1: Definition of Mass

A property that reflects the quantity of matter within a sample is defined as mass

Step 2: Calculations of mass of 2.63 mol of allyl sulfide

Mass of allyl sulfide $\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{63}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{llyl}\mathbf{ }\mathbf{sulfide}\left(\frac{114.23g a \mathrm{llyl} \mathrm{sulfide}}{\mathrm{mol} a \mathrm{llyl} \mathrm{sulfide}}\right)$

Mass of allyl sulfide $\mathbf{=}\mathbf{300}\mathbf{.}\mathbf{42}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{lly}\mathbf{ }\mathbf{l}\mathbf{ }\mathbf{sulfide}$

Moles of allyl sulfide  $\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{7}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{allyl}\mathbf{ }\mathbf{sulfide}\mathbf{\left(}\frac{\mathbf{mol}\mathbf{ }\mathbf{allysulfide}}{\mathbf{114}\mathbf{.}\mathbf{23}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{ }\mathbf{allylsulfide}}\mathbf{\right)}$

Moles of allyl sulfide $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{313}\mathbf{ }\mathbf{m}\mathbf{ }\mathbf{ol}\mathbf{ }\mathbf{ally}\mathbf{ }\mathbf{l}\mathbf{ }\mathbf{sulfide}$

Step 2: Calculations of the number of carbon atoms in 35.7 g of allyl sulfide

The number of C atoms

$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{313}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{ally}\mathbf{ }\mathbf{sulfide}\left(\frac{6 \mathrm{mol} C \mathrm{atoms}}{1\mathrm{mol} \mathrm{allyl} \mathrm{sulfide}}\right)\left(\frac{6.022\times {10}^{23}C \mathrm{atoms}}{\mathrm{molC} \mathrm{atoms} }\right)$

Therefore, the number of C atoms $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{13}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{C}\mathbf{ }\mathbf{atoms}$