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Calculate the mass of carbon in the following compounds. (a) \(1.80 \mathrm{~g} \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) (b) \(0.00223 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (c) \(22.1 \mathrm{~g} \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{~N}\)

Short Answer

Expert verified
(a) 1.17 g, (b) 0.000252 g, (c) 15.6 g

Step by step solution

01

Understand the Problem

We need to calculate the mass of carbon present in different compounds given in grams. To do this, we first need to determine the molar mass of each compound and the percent mass of carbon in each compound.
02

Calculate Molar Masses

For each compound, calculate the molar mass:- **Compound (a)**: \( \text{C}_4\text{H}_{10}\text{O} \) Molar Mass \( = 4\times12.01 + 10\times1.008 + 16.00 = 74.12 \, \text{g/mol} \)- **Compound (b)**: \( \text{Na}_2\text{CO}_3 \) Molar Mass \( = 2\times22.99 + 12.01 + 3\times16.00 = 105.99 \, \text{g/mol} \)- **Compound (c)**: \( \text{C}_5\text{H}_{11}\text{N} \) Molar Mass \( = 5\times12.01 + 11\times1.008 + 14.01 = 85.15 \, \text{g/mol} \)
03

Calculate Percent Mass of Carbon

Next, compute the percent mass of carbon in each compound:- **Compound (a)**: Percent C = \( \frac{4 \times 12.01}{74.12} \times 100 \approx 64.8\% \)- **Compound (b)**: Percent C = \( \frac{12.01}{105.99} \times 100 \approx 11.3\% \)- **Compound (c)**: Percent C = \( \frac{5 \times 12.01}{85.15} \times 100 \approx 70.6\% \)
04

Calculate Mass of Carbon in Given Mass of Molecules

Use the percent mass of carbon calculated earlier to find the mass of carbon in the given mass of each compound:- **Compound (a)**: Mass of \( \text{C} = 1.80 \, \text{g} \times \frac{64.8}{100} = 1.17 \, \text{g} \)- **Compound (b)**: Mass of \( \text{C} = 0.00223 \, \text{g} \times \frac{11.3}{100} = 0.000252 \, \text{g} \)- **Compound (c)**: Mass of \( \text{C} = 22.1 \, \text{g} \times \frac{70.6}{100} = 15.6 \, \text{g} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
When calculating the molar mass of a compound, it is essential to know the elements present and their individual atomic masses. The molar mass is the sum of the masses of all atoms in the compound in grams per mole. Here's the simplified procedure:

- Identify each element in the compound and note the number of moles of each element.
- Multiply the atomic mass of each element by the number of moles in the compound.
- Add all the results together to get the total molar mass.

For example, in compound (a) \(\text{C}_4\text{H}_{10}\text{O}\), the contribution of carbon is considerable. To find the molar mass, compute \(4 \times 12.01\) for carbon, \(10 \times 1.008\) for hydrogen, and \(16.00\) for oxygen. Adding these gives the molar mass: \(74.12 \, \text{g/mol}\).

Understanding molar mass helps us determine how much of a compound's total weight is made up by a particular element, such as carbon.
Percent Composition
Percent composition is vital in chemistry to understand what fraction of a compound's total mass is due to each element. Calculating percent composition involves dividing the mass of each type of atom by the molar mass of the compound and multiplying by 100.

This provides the percentage of the compound by mass that comes from each element. For compound (a) \(\text{C}_4\text{H}_{10}\text{O}\), the percent by mass of carbon is found using:

- Calculate the total mass of carbon atoms in the compound: \(4 \times 12.01\)
- Divide this by the compound's total molar mass \(74.12\)
- Multiply by 100 to get \(64.8\%\)

Thus, 64.8% of the mass of \(\text{C}_4\text{H}_{10}\text{O}\) is due to carbon. This knowledge aids in analyzing the compound's properties and its reactions with other substances.
Stoichiometry
Stoichiometry involves calculations that relate the masses and percent compositions of reactants and products in chemical reactions. It allows us to predict how substances interact and transform during reactions.

To find the mass of carbon in a specific amount of compound, multiply the mass of the compound by the percent composition of carbon. This converts the percentage into a tangible mass.

- Using compound (a), for instance, the percent mass of carbon was previously calculated as \(64.8\%\).
- If you're given \(1.80\, \text{g}\) of \(\text{C}_4\text{H}_{10}\text{O}\), multiply \(1.80\, \text{g}\) by \(\frac{64.8}{100}\) to find \(1.17\, \text{g of C}\).

Understanding stoichiometry is critical for determining how much of each element is involved in reactions, particularly when calculating yields and assessing the efficiency of chemical processes.

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Most popular questions from this chapter

Molecular nitrogen can be converted to NO in two steps. The equations follow. These two reactions are the first steps in the industrially important conversion of nitrogen into nitric acid. Calculate the mass of \(\mathrm{NO}\) formed from \(100 \mathrm{~g} \mathrm{~N}_{2}\) and excess \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\). $$ \begin{array}{l} \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \\ 4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O} \end{array} $$

In compounds called peroxides, the oxygen atoms do not have oxidation numbers common for oxygen atoms. On the basis of the formula of sodium peroxide, \(\mathrm{Na}_{2} \mathrm{O}_{2}\), what is the oxidation number of the \(\mathrm{O}\) atoms?

What is the percentage, by mass, of each element in the following substances? (a) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{~N}_{2}\) (c) \(\mathrm{Fe}_{2} \mathrm{O}_{3}\)

A \(2.770-\mathrm{g}\) sample containing only carbon, hydrogen, and oxygen burns in excess \(\mathrm{O}_{2}\) to produce \(4.06 \mathrm{~g} \mathrm{CO}_{2}\) and \(1.66 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) Calculate the mass of each element in the sample and the mass percentage of each element in the compound.

A \(0.459-\mathrm{g}\) sample that contains only carbon, hydrogen, and oxygen reacts with an excess of \(\mathrm{O}_{2}\) to produce \(0.170 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.0348 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). What is the percentage of \(\mathrm{C}\) and \(\mathrm{H}\) in the starting material?

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