Suppose a substance has been prepared that is composed of carbon, hydrogen, and nitrogen. When 0.1156 gram of this compound is reacted with oxygen, 0.1638 gram of carbon dioxide (CO2) and 0.1676 gram of water (H2O) are collected. Assuming that all the carbon in the compound is converted to CO2

A. Determine the mass of Carbon and hydrogen in the sample (2)
B. Calculate the percent composition of Carbon, Nitrogen and Hydrogen in the substance ( 9)
C. Calculate the mass of nitrogen in the substance (2)
D. Calculate the empirical formula of the sample (3)
Suppose the molecular mass of the substance was 93 g/mol, find the molecular formula. (4)

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To answer the given questions, we need to use stoichiometry and the concept of percent composition.

A. To determine the mass of carbon and hydrogen in the sample:
1. Start by calculating the mass of carbon in the compound:
- From the reaction, we know that 0.1156 grams of the compound produces 0.1638 grams of CO2.
- Since all the carbon in the compound is converted to CO2, the mass of carbon in the sample is also 0.1638 grams.

2. Next, calculate the mass of hydrogen in the compound:
- From the reaction, we know that 0.1156 grams of the compound produces 0.1676 grams of H2O.
- Since water (H2O) consists of 2 hydrogen atoms, the molar mass of 1 mole of H2O is 18 grams.
- Thus, the moles of water produced is 0.1676 grams / 18 grams/mol = 0.0093 moles.
- Since 1 mole of water contains 2 moles of hydrogen, the moles of hydrogen in the sample is 2 * 0.0093 moles = 0.0186 moles.
- To convert moles to grams, we can multiply by the molar mass of hydrogen, which is approximately 1 gram/mol.
- Therefore, the mass of hydrogen in the sample is 0.0186 moles * 1 gram/mol = 0.0186 grams.

Therefore, the mass of carbon in the sample is 0.1638 grams and the mass of hydrogen in the sample is 0.0186 grams.

B. To calculate the percent composition of carbon, nitrogen, and hydrogen in the substance:
1. Calculate the total mass of the substance by summing the masses of carbon, hydrogen, and nitrogen:
- The mass of carbon is 0.1638 grams (from part A).
- The mass of hydrogen is 0.0186 grams (from part A).
- The mass of nitrogen can be calculated by subtracting the mass of carbon and hydrogen from the mass of the sample: 0.1156 grams - 0.1638 grams - 0.0186 grams = 0.1156 grams - 0.1824 grams = 0.0670 grams.
- Therefore, the total mass of the substance is 0.1638 grams + 0.0186 grams + 0.0670 grams = 0.2494 grams.

2. Calculate the percent composition of each element:
- Percent composition of carbon = (mass of carbon / total mass of substance) * 100% = (0.1638 grams / 0.2494 grams) * 100%.
- Percent composition of hydrogen = (mass of hydrogen / total mass of substance) * 100% = (0.0186 grams / 0.2494 grams) * 100%.
- Percent composition of nitrogen = (mass of nitrogen / total mass of substance) * 100% = (0.0670 grams / 0.2494 grams) * 100%.

Therefore, calculate the percentages using the above formulas.

C. To calculate the mass of nitrogen in the substance:
- The mass of nitrogen is 0.0670 grams (from part B).

D. To calculate the empirical formula of the sample:
1. Determine the moles of carbon, hydrogen, and nitrogen:
- Moles of carbon = mass of carbon / molar mass of carbon.
- Moles of hydrogen = mass of hydrogen / molar mass of hydrogen.
- Moles of nitrogen = mass of nitrogen / molar mass of nitrogen.
- Use the periodic table to find the molar masses of each element.

2. Divide the moles of each element by the smallest mole value obtained above.
- This step ensures that the subscripts in the empirical formula represent the simplest whole number ratio of elements.

3. Write the empirical formula using the obtained mole ratios:

For example, if the mole ratios are C:H:N = 1:2:3, the empirical formula would be CH2N3.

D. To find the molecular formula:
1. Determine the empirical formula weight by summing the atomic weights of the elements in the empirical formula.

2. Calculate the ratio of the molecular weight given (93 g/mol) to the empirical formula weight.

3. Multiply each subscript in the empirical formula by the ratio obtained in step 2.

This will give the molecular formula of the substance.

Please note that the empirical formula may or may not be the same as the molecular formula, depending on the above calculations.

Your teacher has kindly laid out the steps A-D in order to solve this. What is your question on this?