Question

chemistry – unit 4 worksheet 4
answer the following questions on your own piece of paper. be sure to show all
mathematical work and reasoning and use complete sentences in explanations.

1. table sugar is a compound known as sucrose. sucrose is composed of the elements

carbon, hydrogen, and oxygen. analysis of a 20.0 g of sucrose from a bag of sugar
finds that the sugar is composed of 8.44 g of carbon, 1.30 g of hydrogen, and 10.26 g
of oxygen.
a. express, as fractions, the ratio of the mass of each element to the total mass
of the sample.
b. using these ratios, calculate the percent composition by mass of each element
in the compound.

1. a similar chemical analysis is performed on a 500.0 g sample of the sugar isolated

from a sample of pure sugar cane. analysis shows this sample contains 211.0 g of
carbon, 32.5 g of hydrogen, and 256.5 g of oxygen.
a. determine the percent composition by mass of each element in the sugar cane
sample.
b. could the sugar in this sample be sucrose? justify your conclusion.

1. a similar chemical analysis is performed on a 200.0g sample of the sugar found in

corn syrup. this sample contains 80.0g of carbon, 13.3 g of hydrogen and 106.7 g of
oxygen.
a. determine the percent composition by mass of each element in the sugar cane
sample.
b. could the sugar in corn syrup be sucrose? justify your conclusion.

1. a 1.0 g sample of hydrogen reacts completely with 19.0 g of fluorine to form a

compound of hydrogen and fluorine.
a. what is the percent by mass of each element in the compound?
b. what mass of hydrogen would be present in a 50 g sample of this compound?
c. justify your answer to b.

1. explain how the previous examples help to illustrate the law of definite

proportions.

Explanation:

Step1: Solve 1a - Mass Ratios for Sucrose

Total mass of sucrose sample = \( 8.44 + 1.30 + 10.26 = 20.0 \, \text{g} \) (check: \( 8.44 + 1.30 = 9.74 \); \( 9.74 + 10.26 = 20.0 \)).

• Carbon ratio: \( \frac{8.44}{20.0} \)
• Hydrogen ratio: \( \frac{1.30}{20.0} \)
• Oxygen ratio: \( \frac{10.26}{20.0} \)

Step2: Solve 1b - Percent Composition for Sucrose

Percent composition = (mass ratio) \( \times 100\% \).

• Carbon: \( \frac{8.44}{20.0} \times 100\% = 42.2\% \)
• Hydrogen: \( \frac{1.30}{20.0} \times 100\% = 6.50\% \)
• Oxygen: \( \frac{10.26}{20.0} \times 100\% = 51.3\% \)

Step3: Solve 2a - Percent Composition for Sugar Cane

Total mass = \( 211.0 + 32.5 + 256.5 = 500.0 \, \text{g} \) (check: \( 211.0 + 32.5 = 243.5 \); \( 243.5 + 256.5 = 500.0 \)).

• Carbon: \( \frac{211.0}{500.0} \times 100\% = 42.2\% \)
• Hydrogen: \( \frac{32.5}{500.0} \times 100\% = 6.50\% \)
• Oxygen: \( \frac{256.5}{500.0} \times 100\% = 51.3\% \)

Step4: Solve 2b - Is Sugar Cane Sucrose?

Compare percent compositions. Sugar cane sample has same % composition as sucrose (42.2% C, 6.50% H, 51.3% O). By Law of Definite Proportions, compounds have fixed element ratios, so it could be sucrose.

Step5: Solve 3a - Percent Composition for Corn Syrup

Total mass = \( 80.0 + 13.3 + 106.7 = 200.0 \, \text{g} \) (check: \( 80.0 + 13.3 = 93.3 \); \( 93.3 + 106.7 = 200.0 \)).

• Carbon: \( \frac{80.0}{200.0} \times 100\% = 40.0\% \)
• Hydrogen: \( \frac{13.3}{200.0} \times 100\% = 6.65\% \)
• Oxygen: \( \frac{106.7}{200.0} \times 100\% = 53.35\% \)

Step6: Solve 3b - Is Corn Syrup Sucrose?

Corn syrup has different % composition (40.0% C vs 42.2% C, 6.65% H vs 6.50% H, 53.35% O vs 51.3% O) than sucrose. Thus, it is not sucrose.

Step7: Solve 4a - Percent Composition for H-F Compound

Total mass = \( 1.0 + 19.0 = 20.0 \, \text{g} \).

• Hydrogen: \( \frac{1.0}{20.0} \times 100\% = 5.0\% \)
• Fluorine: \( \frac{19.0}{20.0} \times 100\% = 95.0\% \)

Step8: Solve 4b - Mass of H in 50 g Sample

Mass of H = (percent H) \( \times \) sample mass \( \div 100\% \).
\( 5.0\% \times 50 \, \text{g} = 2.5 \, \text{g} \)

Step9: Solve 4c - Justify 4b

By Law of Definite Proportions, the compound’s element ratio is fixed. So 5.0% of any sample mass is H. For 50 g, \( 0.05 \times 50 = 2.5 \, \text{g} \).

Step10: Solve 5 - Law of Definite Proportions

Sucrose and sugar cane sample have identical % composition (fixed proportions), showing a compound’s elements are in fixed ratio. Corn syrup differs, so it’s a different compound. H-F compound has fixed 5% H/95% F, regardless of sample size (e.g., 50 g sample has 2.5 g H). These examples show compounds have definite, constant element proportions (Law of Definite Proportions).

Answer:

1a

• Carbon: \( \boldsymbol{\frac{8.44}{20.0}} \)
• Hydrogen: \( \boldsymbol{\frac{1.30}{20.0}} \)
• Oxygen: \( \boldsymbol{\frac{10.26}{20.0}} \)

1b

• Carbon: \( \boldsymbol{42.2\%} \)
• Hydrogen: \( \boldsymbol{6.50\%} \)
• Oxygen: \( \boldsymbol{51.3\%} \)

2a

• Carbon: \( \boldsymbol{42.2\%} \)
• Hydrogen: \( \boldsymbol{6.50\%} \)
• Oxygen: \( \boldsymbol{51.3\%} \)

2b

Yes, the sugar cane sample could be sucrose. Its percent composition (42.2% C, 6.50% H, 51.3% O) matches sucrose’s, consistent with the Law of Definite Proportions (compounds have fixed element ratios).

3a

• Carbon: \( \boldsymbol{40.0\%} \)
• Hydrogen: \( \boldsymbol{6.65\%} \)
• Oxygen: \( \boldsymbol{53.35\%} \)

3b

No, the corn syrup sugar is not sucrose. Its percent composition (40.0% C, 6.65% H, 53.35% O) differs from sucrose’s, so it has a different element ratio (violates fixed proportions for sucrose).

4a

• Hydrogen: \( \boldsymbol{5.0\%} \)
• Fluorine: \( \boldsymbol{95.0\%} \)

4b

\( \boldsymbol{2.5 \, \text{g}} \)

4c

The compound’s percent composition is fixed (5.0% H) by the Law of Definite Proportions. Thus, 5.0% of 50 g is hydrogen: \( 0.05 \times 50 = 2.5 \, \text{g} \).

5

The examples illustrate the Law of Definite Proportions (compounds have fixed element mass ratios):

• Sucrose and sugar cane sample have identical % composition (fixed C:H:O), confirming constant proportions.
• Corn syrup has a different % composition, so it is a different compound.
• The H-F compound has a fixed 5% H/95% F ratio, with the same proportion (2.5 g H in 50 g) regardless of sample size.

These show compounds contain elements in definite, constant proportions.