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.

Question

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.

Naomi Aronson · Accepted Answer

Take the product...#"molar quantity"xx"molar mass"#; I gets a tad over #1*g#...

You got the molar quantity...#4.52xx10^-3*mol#...(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#

Caleb Adams · Answer

The mass of #4.52xx"10"^(-3) "mol C"_20"H"_42"# is #"1.27 g"#.

If the mass of #4.52xx"10"^(-3) "mol C"_20"H"_42"# is what you are looking for: Determine the molar mass of #"C"_20"H"_42"#. #"Molar mass"##color(white)(.)## "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 #4.52xx"10"^(-3) "mol C"_20"H"_42"# Multiply mol #"C"_20"H"_42"# by its molar mass. #4.52xx"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

Cooper Anderson · Answer

undefined To calculate the mass in grams of $4.52 \times 10^{-3}$ $C_{20}H_{42}$, use the molar masses provided:

Molar mass of carbon ($C$) = 12.0 g/mol
Molar mass of hydrogen ($H$) = 1.0 g/mol

First, calculate the molar mass of $C_{20}H_{42}$:
$$
(20 \times 12.0 \, \text{g/mol}) + (42 \times 1.0 \, \text{g/mol}) = 240.0 \, \text{g/mol} + 42.0 \, \text{g/mol} = 282.0 \, \text{g/mol}
$$

Next, multiply the molar mass by the given amount in moles:
$$
4.52 \times 10^{-3} \, \text{mol} \times 282.0 \, \text{g/mol} = 1.27704 \, \text{g}
$$

Therefore, the mass of $4.52 \times 10^{-3}$ $C_{20}H_{42}$ is approximately $1.28$ grams.