What ratio is used to carry out each conversion? a. mol CH4 to grams CH4 b. L CH4 (g) to mol CH4 (g) (at STP) c. molecules CH4 to mol CH4 I honestly don't kno what it's asking me I got 1 mol CH4 / 16 (g) CH4 for a. 1 mol/22.4 L for b. and 6.02 x 10^23 molecules/1 mol but I don't think this is right, is it?

Question

Trace Green

Chemistry

6 January, 09:16

What ratio is used to carry out each conversion? a. mol CH4 to grams CH4 b. L CH4 (g) to mol CH4 (g) (at STP) c. molecules CH4 to mol CH4 I honestly don't kno what it's asking me I got 1 mol CH4 / 16 (g) CH4 for a. 1 mol/22.4 L for b. and 6.02 x 10^23 molecules/1 mol but I don't think this is right, is it?

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Answers (1)

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Shane · Answer 1

a. mol CH4 to grams CH4

To convert from moles to grams, we need the molar mass of CH4 which is 16.05 g/mol.

b. L CH4 (g) to mol CH4 (g) (at STP)

To convert from L to mol CH4, we need the relation 1 mol of an ideal gas is equal to 22.4 L of that gas. The ratio would be 1/22.4 mol/L.

c. molecules CH4 to mol CH4

To convert from molecules to mol, we need the avogadro's number which is 6.022x10^23 units / 1 mol. The ratio to be multiplied would be 1/6.022x10^23 mol/molecules.