Answer:
The empirical formula is =
Explanation:
Mass of water obtained = ![6.954\times 10^{-2}\ mg](https://tex.z-dn.net/?f=6.954%5Ctimes%2010%5E%7B-2%7D%5C%20mg)
Also, 1 mg =
g
So, mass of water = ![6.954\times 10^{-5}\ g](https://tex.z-dn.net/?f=6.954%5Ctimes%2010%5E%7B-5%7D%5C%20g)
Molar mass of water = 18 g/mol
Moles of
=
/18 g/mol = 3.8634×10⁻⁶ moles
2 moles of hydrogen atoms are present in 1 mole of water. So,
<u>Moles of H = 2 x 3.8634×10⁻⁶ = 7.7267×10⁻⁶ moles
</u>
Molar mass of H atom = 1.008 g/mol
<u>Mass of H in molecule = 7.7267×10⁻⁶ x 1.008 = 7.6653×10⁻⁶ g
</u>
Mass of carbon dioxide obtained = 0.1697 mg
Also, 1 mg =
g
So, mass of carbon dioxide = ![1.697\times 10^{-4}\ g](https://tex.z-dn.net/?f=1.697%5Ctimes%2010%5E%7B-4%7D%5C%20g)
Molar mass of carbon dioxide = 44.01 g/mol
<u>Moles of
=
/44.01 g/mol = 3.856×10⁻⁶ moles
</u>
1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,
Moles of C = 3.856×10⁻⁶ moles
Molar mass of C atom = 12.0107 g/mol
<u>Mass of C in molecule = 3.856×10⁻⁶ x 12.0107 = 46.3132 ×10⁻⁶ g</u>
Given that the Menthol only contains hydrogen, oxygen and carbon. So,
Mass of O in the sample = Total mass - Mass of C - Mass of H
Mass of the sample = ![6.030\times 10^{-2}\ mg](https://tex.z-dn.net/?f=6.030%5Ctimes%2010%5E%7B-2%7D%5C%20mg)
Also, 1 mg =
g
So, mass of sample = ![60.30\times 10^{-6}\ g](https://tex.z-dn.net/?f=60.30%5Ctimes%2010%5E%7B-6%7D%5C%20g)
<u>Mass of O in sample = ( 60.30 - 7.6653 - 46.3132 ) ×10⁻⁶ g= 6.3215×10⁻⁶ g</u>
Molar mass of O = 15.999 g/mol
<u>Moles of O = 6.3215×10⁻⁶ / 15.999 = 0.3951×10⁻⁶ moles
</u>
Taking the simplest ratio for H, O and C as:
7.7267×10⁻⁶ : 0.3951×10⁻⁶ : 3.856×10⁻⁶
= 20 : 1 : 10
<u>The empirical formula is =
</u>