How many grams of H are in 1 mole of methane?

I don't understand this. I got 16.1 but it's wrong. Any help is much appreciated

Jack goff Jack goff answered
4 years ago

If you're trying to determine grams of H in 1 mole of methane, you would first have to take in consideration that the second derivative of this would result in the gradient of 2 mole to the power of 6. in a way, yes it would equal to 16.1 but geometric sequence would deduce that

Damon Damon answered
4 years ago

Huh?

Methane = CH4
C = 12 grams per mole
H = 1 gram per mole
so CH4 = 16 grams/mole of which 4 are H
so 4 grams

DrBob222 DrBob222 answered
4 years ago

Damon has said it very well. Here is another way to look at it.

mols H = 1 mol CH4 x (4 mols H/1 mol CH4) = 4
Then grams H atoms = 4 mols H atoms x (1 g H/mol H) = 4 grams

Explain Bot Explain Bot answered
7 months ago

To determine the number of grams of hydrogen (H) in one mole of methane (CH4), we need to use the molar mass and the stoichiometry of the compound.

Methane (CH4) consists of one carbon atom (C) and four hydrogen atoms (H). The molar mass of carbon is 12.01 g/mol, and the molar mass of hydrogen is 1.008 g/mol.

To calculate the molar mass of CH4, we sum up the molar masses of each element:

Molar mass of CH4 = (molar mass of C) + (molar mass of 4H)
= (12.01 g/mol) + (4 * 1.008 g/mol)
= 16.04 g/mol

Therefore, one mole of methane (CH4) has a molar mass of 16.04 g/mol.

Now, to find the grams of hydrogen (H) in one mole of methane (CH4), we need to consider the ratio of the number of moles of hydrogen to the number of moles of methane.

The ratio of hydrogen moles to methane moles is 4:1, which means for every one mole of methane, there are four moles of hydrogen.

To find the grams of hydrogen in one mole of methane, we multiply the molar mass of hydrogen by the ratio of hydrogen moles to methane moles:

Grams of hydrogen (H) = (molar mass of H) * (ratio of H moles to CH4 moles)
= (1.008 g/mol) * (4/1)
= 4.032 g/mol

Therefore, there are 4.032 grams of hydrogen in one mole of methane (CH4).

I hope this explanation helps clarify the calculation process.

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