Answer:
Explanation:
<u>1) Content of C:</u>
All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.
- Mass of C in 1.813 mg of CO₂
Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:
12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂
⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C
number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol
<u>2) Content of H</u>
All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O
- Mass of H in 0.4639 mg of H₂O
Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:
2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O
⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H
number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol
<u>3) Content of N</u>
All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂
- Mass of N in 0.2885 mg of N₂ is 0.2885 mg
number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol
<u>4) Content of O</u>
The mass of O is calculated by difference:
- Mass of O = mass of sample - mass of C - mass of H - mass of N
Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N
Mass of O = 0.1648 mg
- Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol
<u>5) Ratios</u>
Divide every number of mililmoles by the smallest number of milimoles:
- C: 0.041195 / 0.01030 = 4
- H: 0.051501 / 0.01030 = 5
- N: 0.020597 / 0.01030 = 2
<u>6) Empirical formula:</u>
<u>7) Calculate the approximate mass of the empirical formula:</u>
- 4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol
So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.
Thus, the estimate is 194 g/mol
