Chapter 10: Problem 215
Hydrazine contains 87.45% nitrogen and 12.55% hydrogen, and has a molar mass of 32.04 g/mol. Determine hydrazine’s molecular formula. Record the molecular formula in Table 10.6.
Short Answer
Expert verified
The molecular formula of hydrazine is N2H4.
Step by step solution
01
Determine the moles of each element in a 100 g sample
First, we need to convert the percentage composition to quantity of each element in grams. Assuming we have a 100 g sample, we would have 87.45 g of nitrogen (N) and 12.55 g of hydrogen (H).
Now, we will convert these amounts into moles using the molar masses of the elements. The molar mass of nitrogen is 14.01 g/mol, and the molar mass of hydrogen is 1.01 g/mol.
Moles of nitrogen (N) = \( \frac{87.45 \;\text{g}}{14.01 \;\text{g/mol}} \) = 6.24 mol
Moles of hydrogen (H) = \( \frac{12.55 \;\text{g}}{1.01 \;\text{g/mol}} \) = 12.42 mol
02
Determine mole ratio of elements
Next, we need to determine the mole ratio between nitrogen and hydrogen. To do this, divide the moles of each element by the smallest number of moles:
Mole ratio of N:H = \( \frac{6.24 \;\text{mol}}{6.24 \;\text{mol}} \) : \( \frac{12.42 \;\text{mol}}{6.24 \;\text{mol}} \) = 1:2
This means that for every 1 nitrogen atom, there are 2 hydrogen atoms in the molecule.
03
Determine the molecular formula using the molar mass
The empirical formula we obtained in the previous step is NH2. Now, we need to determine the molecular formula using the given molar mass of hydrazine, 32.04 g/mol.
First, find the molar mass of the empirical formula (NH2):
Molar mass = (1 × 14.01) + (2 × 1.01) = 16.03 g/mol
Now, divide the molar mass of hydrazine by the molar mass of the empirical formula:
Molecular formula multiplier = \( \frac{32.04 \;\text{g/mol}}{16.03 \;\text{g/mol}} \) = 2
Multiply the empirical formula by the molecular formula multiplier to get the molecular formula:
Molecular formula = 2 × (NH2) = N2H4
Therefore, the molecular formula of hydrazine is N2H4.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Percentage Composition
When solving chemical problems, knowing the percentage composition of a compound is valuable. It tells us how much of each element is present in a compound.
For hydrazine, these percentages are given as 87.45% nitrogen and 12.55% hydrogen. By imagining a 100 g sample, we can directly translate these percentages into grams.
This gives us 87.45 g of nitrogen and 12.55 g of hydrogen.
For hydrazine, these percentages are given as 87.45% nitrogen and 12.55% hydrogen. By imagining a 100 g sample, we can directly translate these percentages into grams.
This gives us 87.45 g of nitrogen and 12.55 g of hydrogen.
- Percentage composition helps us find out the relative amounts of each element in a substance, using a simple percentage-to-gram conversion.
- This forms the foundation for further calculations, such as finding the number of moles or determining the empirical formula.
Molar Mass Calculation
Calculating the molar mass is a crucial step in chemistry, as it bridges the gap between the microscopic world of atoms and the macroscopic world of grams.
Each element has a known molar mass, which is the mass of one mole of its atoms, expressed in grams per mole (g/mol). For our elements:
The calculations are as follows: - Moles of N = \( \frac{87.45 \, \text{g}}{14.01 \, \text{g/mol}} \) = 6.24 moles. - Moles of H = \( \frac{12.55 \, \text{g}}{1.01 \, \text{g/mol}} \) = 12.42 moles. Understanding molar mass is essential, as it is used to convert mass measurements to mole numbers, the fundamental unit of chemical quantity.
Each element has a known molar mass, which is the mass of one mole of its atoms, expressed in grams per mole (g/mol). For our elements:
- Nitrogen (N) has a molar mass of 14.01 g/mol.
- Hydrogen (H) has a molar mass of 1.01 g/mol.
The calculations are as follows: - Moles of N = \( \frac{87.45 \, \text{g}}{14.01 \, \text{g/mol}} \) = 6.24 moles. - Moles of H = \( \frac{12.55 \, \text{g}}{1.01 \, \text{g/mol}} \) = 12.42 moles. Understanding molar mass is essential, as it is used to convert mass measurements to mole numbers, the fundamental unit of chemical quantity.
Empirical Formula
An empirical formula represents the simplest whole-number ratio of atoms in a compound. It doesn't tell us exactly how many atoms are in a molecule, but it provides a basic framework.
For hydrazine, after calculating the moles, we use the smallest mole value to find the ratio: - Mole ratio of N:H is \( \frac{6.24}{6.24} \):\( \frac{12.42}{6.24} \) = 1:2.
This indicates that for every nitrogen atom, there are two hydrogen atoms, giving us the empirical formula NH2.
For hydrazine, after calculating the moles, we use the smallest mole value to find the ratio: - Mole ratio of N:H is \( \frac{6.24}{6.24} \):\( \frac{12.42}{6.24} \) = 1:2.
This indicates that for every nitrogen atom, there are two hydrogen atoms, giving us the empirical formula NH2.
- Finding the empirical formula involves using mole ratios and often requires rounding to the nearest whole number.
- It's an important step before calculating the actual molecular formula of a compound.
Mole Ratio Calculation
Mole ratio calculation is fundamental to translating the empirical formula to the molecular formula. It involves determining the relationship between the actual molar mass and the molar mass of the empirical formula.
In hydrazine's case, the empirical formula NH2 has a molar mass of 16.03 g/mol. The given molar mass of hydrazine is 32.04 g/mol.
To determine the molecular formula, divide the molar mass of hydrazine by the molar mass of the empirical formula: - Molecular formula multiplier = \( \frac{32.04 \, \text{g/mol}}{16.03 \, \text{g/mol}} \) = 2.
This multiplier tells us how many times to scale the empirical formula to match the actual molar mass.
In hydrazine's case, the empirical formula NH2 has a molar mass of 16.03 g/mol. The given molar mass of hydrazine is 32.04 g/mol.
To determine the molecular formula, divide the molar mass of hydrazine by the molar mass of the empirical formula: - Molecular formula multiplier = \( \frac{32.04 \, \text{g/mol}}{16.03 \, \text{g/mol}} \) = 2.
This multiplier tells us how many times to scale the empirical formula to match the actual molar mass.
- The molecular formula multiplier helps calculate the 'real' number of atoms in each molecule.
- In this case, multiplying NH2 by 2 gives us the molecular formula N2H4.