There are ##3.8 × 10^23″ C atoms”## in 10 g of ##”CH₄”## and ##8.2 × 10^23″ C atoms”## in 20 g of ##”C”_3″H”_8##.
##”C”## atoms in ##”CH”_4##
You must convert
##”grams of CH”_4 → “moles of CH”_4 → “molecules of CH”_4 → “atoms of C”##
The molar mass of ##”CH”_4″## is 16.04 g/mol.
You use this to convert grams to moles.
##10 cancel(“g CH”_4) × (1 cancel(“mol CH₄”))/(16.04 cancel(“g CH₄”)) × (6.022 × 10^23 cancel(“molecules CH₄”))/(1 cancel(“mol CH₄”)) × “1 atom C”/(1 cancel(“molecule CH₄”)) = 3.8 × 10^23″atoms C”##
C atoms in ##”C”_3″H”_8##
The molar mass of ##”C”_3″H”_8## is 44.10 g/mol.
Can you show there are ##8.2 × 10^23″ C”## atoms in 20 g of ##”C”_3″H”_8##?
