Question

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?

Answer 1

Below in bold.

Step-by-step 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).

Answer 2

C atoms = 9.6 *10^23 atoms

H atoms = 2.9*10^24 atoms

O atoms = 4.8 *10^23 atoms

Step-by-step 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