7. how many grams of carbon are in 15.0g of co₂?\n8. how many grams of hydrogen are in 12.0g of h₂o?\n9…

7. How many grams of carbon are in 15.0g of CO₂?

Explanation:

Step1: Calculate molar mass of CO₂

The molar mass of C is 12.01 g/mol and O is 16.00 g/mol. So molar mass of CO₂, $M_{CO_2}=12.01 + 2\times16.00=44.01$ g/mol

Step2: Calculate mass of carbon

The mass - percent of C in CO₂ is $\frac{12.01}{44.01}\times100%$. Mass of C in 15.0 g of CO₂, $m_C = 15.0\times\frac{12.01}{44.01}\approx4.09$ g

Answer:

4.09 g

8. How many grams of hydrogen are in 12.0g of H₂O?

Explanation:

Step1: Calculate molar mass of H₂O

The molar mass of H is 1.01 g/mol and O is 16.00 g/mol. So molar mass of H₂O, $M_{H_2O}=2\times1.01+16.00 = 18.02$ g/mol

Step2: Calculate mass of hydrogen

The mass - percent of H in H₂O is $\frac{2\times1.01}{18.02}\times100%$. Mass of H in 12.0 g of H₂O, $m_H=12.0\times\frac{2\times1.01}{18.02} = 1.34$ g

Answer:

1.34 g

9. What is the empirical formula of a compound that is made up of 24.5g of nitrogen and 70.0g of oxygen?

Explanation:

Step1: Calculate moles of N and O

Moles of N, $n_N=\frac{24.5}{14.01}\approx1.75$ mol. Moles of O, $n_O=\frac{70.0}{16.00}=4.375$ mol

Step2: Find mole - ratio

Divide each number of moles by the smaller number of moles. $\frac{n_N}{n_N}=1$, $\frac{n_O}{n_N}=\frac{4.375}{1.75}=2.5$. Multiply by 2 to get whole - numbers. The ratio of N:O is 2:5

Answer:

N₂O₅

10. The elemental mass percent composition of ascorbic acid is 40.92% C, 4.58% H, and 54.5% O. Determine the empirical formula for this compound.

Explanation:

Step1: Assume 100 g of the compound

So we have 40.92 g of C, 4.58 g of H and 54.5 g of O.

Step2: Calculate moles of each element

Moles of C, $n_C=\frac{40.92}{12.01}\approx3.41$ mol. Moles of H, $n_H=\frac{4.58}{1.01}\approx4.53$ mol. Moles of O, $n_O=\frac{54.5}{16.00}\approx3.41$ mol

Step3: Find mole - ratio

$\frac{n_C}{n_C}=1$, $\frac{n_H}{n_C}=\frac{4.53}{3.41}\approx1.33$, $\frac{n_O}{n_C}=1$. Multiply by 3 to get whole - numbers. The ratio of C:H:O is 3:4:3

Answer:

C₃H₄O₃

11. Upon combustion, a 0.8233 - g sample of a compound containing only carbon, hydrogen, and oxygen produces 2.445g CO₂ and 0.6003g H₂O. Determine the empirical formula of the compound.

Explanation:

Step1: Calculate moles of C and H from products

Moles of C from CO₂: $n_C=\frac{2.445}{44.01}\times1 = 0.0556$ mol. Moles of H from H₂O: $n_H=\frac{0.6003}{18.02}\times2=0.0666$ mol

Step2: Calculate mass of C and H

Mass of C, $m_C = 0.0556\times12.01=0.668$ g. Mass of H, $m_H=0.0666\times1.01 = 0.0673$ g

Step3: Calculate mass and moles of O

Mass of O, $m_O=0.8233-(0.668 + 0.0673)=0.088$ g. Moles of O, $n_O=\frac{0.088}{16.00}=0.0055$ mol

Step4: Find mole - ratio

$\frac{n_C}{n_O}=\frac{0.0556}{0.0055}\approx10$, $\frac{n_H}{n_O}=\frac{0.0666}{0.0055}\approx12$, $\frac{n_O}{n_O}=1$

Answer:

C₁₀H₁₂O