Asked by CJ
What is the empirical formula for a molecule containing 52.20 grams manganese (Mn0 and 3.80 grams of Carbon (c)?
Answers
Answered by
DrBob222
mols Mn = grams/atomic mass
mols C = grams/atomic mass
Now find the ratio to each other with the smallest being 1.00. The easy way to do that is to divide the smaller number by itself (which makes it 1.00), the divide the other number by the same small number. Round to whole numbers.
Post your work if you get stuck.
mols C = grams/atomic mass
Now find the ratio to each other with the smallest being 1.00. The easy way to do that is to divide the smaller number by itself (which makes it 1.00), the divide the other number by the same small number. Round to whole numbers.
Post your work if you get stuck.
Answered by
Aaliyah
Mn3C
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