A sample of ethanol has a mass of 36.7 g.

a) How many carbon atoms does the sample contain?

b) How many hydrogen atoms are present?

c) How many oxygen atoms are present?

Below in bold.

Explanation:

1 mole of an element contains 6.022 * 10^23 atoms.

Ethanol has the molecular formula C2H5OH.

So we have 2 carbon atoms, 6 hydrogen atoms ans 1 oxygen atom in the molecule.

Using molar masses the mass of carbon in 36.7 g of C2H5OH

= [ (2 * 12.011) / (2*12.011 + 6*1.008 + 15.999) ] * 36.7

= [ (2 * 12.011) / 46.069] * 36.7

= 19.14 g

So the number of carbon atoms = (19.14 / 12.011) * 6.022 * 10^23

= 9.596 * 10^23. (Answer).

Using molar masses the mass of hydrogen in 36.7 g of C2H5OH

= [ (6 * 1.008) / (46.069) ] * 36.7

= 4.818 g

So the number of hydrogen atoms = (4.818 / 1.008) * 6.022 * 10^23

= 2.878 * 10^24 (Answer).

Using molar masses the mass of oxygen in 36.7 g of C2H5OH

= [ (15.999) / (46.069) ] * 36.7

= 12.745 g

So the number of hydrogen atoms = (12.745 / 15.999) * 6.022 * 10^23

= 2.924 * 10^24 (Answer).

C atoms = 9.6 * 10^23 atoms

H atoms = 2.9*10^24 atoms

O atoms = 4.8 * 10^23 atoms

Explanation:

Step 1: Data given

Mass of ethanol = 36.7 grams

Molar mass of ethanol = 46.07 g/mol

Step 2: Calculate moles of ethanol

Moles ethanol = mass / molar mass

Moles ethanol = 36.7 grams / 46.07 g/mol

Moles ethanol = 0.797 moles

Step 3: Calculate moles

For 1 mol C2H5OH we have 2 moles C, 6 moles H and 1 mol O

For 0.797 moles C2H5OH we have:

2*0.797 = 1.594 moles C

6*0.797 = 4.782 moles H

1*0.797 = 0.797 moles O

Step 4: Calculate carbon atoms

C atoms = moles C * number of Avogadro

C atoms = 1.594 moles * 6.02 * 10^23

C atoms = 9.6 * 10^23 atoms

Step 5: Calculate H atoms

H atoms = 4.782 moles * 6.02 * 10^23

H atoms = 2.9*10^24 atoms

Step 6: Calculate O atoms

O atoms = 0.797 moles * 6.02 * 10^23

O atoms = 4.8 * 10^23 atoms