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Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol i

Ket [755]

Answer:

Empirical Formula = C₁₀H₂₀O

Solution:

Data Given:

Mass of Menthol = 9.045 × 10⁻² mg = 9.045 × 10⁻⁵ g

Mass of CO₂ = 0.2546 mg = 0.0002546 g

Mass of H₂O = 0.1043 mg = 0.0001043 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100

%C = (2.814) × (12 ÷ 44) × 100

%C = 2.814 × 0.2727 × 100

%C = 76.73 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100

%H = (1.153) × (2.02 ÷ 18.02) × 100

%H = 1.153 × 0.1120 × 100

%H = 12.91 %

%O = 100% - (%C + %H)

%O = 100% - (76.73% + 12.91%)

%O = 100% - 89.64%

%O = 10.36 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 76.73 ÷ 12.01

Moles of C = 6.3888 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 12.91 ÷ 1.01

Moles of H = 12.7821 mol

Moles of O = %O ÷ At.Mass of O

Moles of O = 10.36 ÷ 16.0

Moles of O = 0.6475 mol

Step 3: Find out mole ratio and simplify it;

C H O

6.3888 12.7821 0.6475

6.3888/0.6475 12.7821/0.6475 0.6475/0.6475

9.86 19.74 1

≈ 10 ≈ 20 1

Result:

Empirical Formula = C₁₀H₂₀O₁

8 0

3 years ago

How much of a radioactive parent isotope will remain after three half-lives have passed?

Leona [35]

After 1 half life, 1/2
After 2 half lives, 1/4
After 3 half lives, 1/8

6 0

3 years ago

Which of the following is the balanced equation for the

NNADVOKAT [17]

A.) is the answer =)

6 0

3 years ago

1. What is a taste bud?

tankabanditka [31]

O tissues containing nerve cells for taste

8 0

2 years ago

What is the mass of 2.70 ×10^22 molecules of NaOH (Molar mass = 40.0 g/mol)?

Viefleur [7K]

Data:
Molar Mass of NaOH = 40 g/mol

Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules

6.02*10²³y = 0.270*10²³
$y = \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}$
$\boxed{y \approx 0.04\:mol}$

Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH

$x = 40*0.04$
$\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark$

Answer:
The mass is 1.6 grams

6 0

3 years ago