# How would you the mass in grams of 4.52 xx10^-3 "C"_20"H"_42"? Use the molar masses: C = 12.0 g/mole; H = 1.0 g/mol.

Dec 23, 2017

Take the product...$\text{molar quantity"xx"molar mass}$; I gets a tad over $1 \cdot g$...

#### Explanation:

You got the molar quantity...$4.52 \times {10}^{-} 3 \cdot m o l$...(you have not specified the units in your question, but I assume that these units pertain); and multiply these units by the molar mass...

4.52xx10^-3*molxx(20xx12_("atomic mass of carbon")+42xx1_("atomic mass of hydrogen"))*g*mol^-1=??*g

Dec 23, 2017

The mass of $4.52 \times \text{10"^(-3) "mol C"_20"H"_42}$ is $\text{1.27 g}$.

#### Explanation:

If the mass of $4.52 \times \text{10"^(-3) "mol C"_20"H"_42}$ is what you are looking for:

Determine the molar mass of $\text{C"_20"H"_42}$.

$\text{Molar mass}$$\textcolor{w h i t e}{.}$${\text{C"_20"H}}_{42} :$(20xx"12.0 g/mol C")+(42xx"1.0 g/mol H")="282 g/mol C"_20"H"_42"

Mass of $4.52 \times \text{10"^(-3) "mol C"_20"H"_42}$

Multiply mol $\text{C"_20"H"_42}$ by its molar mass.

$4.52 \times \text{10"^(-3) color(red)cancel(color(black)("mol C"_20"H"_42))xx(282"g C"_20"H"_42)/(1color(red)cancel(color(black)("mol C"_20"H"_42)))="1.27 g C"_20"H"_42}$ rounded to three significant figures