could you help me with this chem problem?

Question

"Determine the number of moles of C in the following subtances"
(i got)
a. 3.8 mol CH4 (2.8 mol C)
b. 0.273 mol C2H6 (.218)
c. 4.89 mol C4H6 (4.04)
e. 22.9 mol C8H18 (19.9)

i just need you all to check my answers... and a definite way to do the problem...

Answer 1

a. has 3.8 mol of C
b. has 0.546 mol of C
c. has 19.56 mol of C
d. has 183.2 mol of C

Look at the formula, then determine how many Carbon atoms are included per formula, then multiply the number of Carbons with the mol of the formula to find the mol of C.

For example, methane (CH4) only has one carbon per CH4. Thus mol of C and mol of CH4 are equal.

Ethane (C2H6) however, has 2 carbons per each molecule of ethane, thus multiply the mol of ethane by 2.

Same concept for the rest, butane, has 4 carbons per formula, so multiply the mol of butane by 4 to determine the mol of C included in butane.

Octane, multiply by 8, because it has 8 carbon atoms per each molecule of C8H18.

Answer 2

ANSWERS

a.) 3.8 moles C
b.) .546 moles C
c.) 19.56 moles C
d.) 183.2 moles C

EXPLANATION

For every molecule the number of atoms of a certain element in the molecule coincides with the subscript that follows its symbol. For instance, in one molecule of CH4 there is 1carbon atom and 4 hydrogen atoms.

A mole, like a dozen, is simply a ratio that stands for a number. Instead of 12 that the word dozen represents, the word mole represents approximately 6.022x(10^23). Therefore when a certain number of moles are applied to a chemical equation, there are that many moles of each atom in the chemical equation multiplied by the subscript.

THE EXAMPLES

a.) Moles H in 3.8 mol CH4? Answer=3.8x4 or 15.2 moles H
a.)Moles C in 3.8 mol CH4? Answer=3.8x1 or 3.8 moles C

b.)Moles H in .273 mol C2H6? Answer=.273x6 or 1.64 moles H
b.)Moles C in .273 mol C2H6? Answer=.273x2 or .546 moles C

Answer 3

Hmm.... I'm not sure... I have forgotten about these...

a. 3.8 mol C
b. 0.273 mol C
c. 4.89 mol C
d. 22.9 mol C

First, I calculated the weight of the carbon...
Then, I converted it into mole...
Hope it helps...