Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App Chapter 7: Problem 82
If \(10.0 \mathrm{~g}\) of an unknown compound composed of carbon, hydrogen, and nitrogen contains \(17.7 \% \mathrm{~N}\) and \(3.8 \times 10^{23}\) atoms of hydrogen, what is its empirical formula?
Short Answer
Expert verified
The empirical formula is \( \text{C}_{5}\text{H}_{5}\text{N} \).
Step by step solution
01
- Calculate mass of nitrogen
Given that the compound has 17.7% nitrogen, the first step is to find the mass of nitrogen in the 10.0 g sample. Mass of nitrogen = 17.7% of 10.0 g = \[ \frac{17.7}{100} \times 10.0 \text{ g} = 1.77 \text{ g} \text{ N} \]
02
- Calculate moles of nitrogen
Using the molar mass of nitrogen (14.01 g/mol), convert the mass of nitrogen calculated in Step 1 into moles. \[ \text{Moles of N} = \frac{1.77 \text{ g}}{14.01 \text{ g/mol}} \ \text{Moles of N} \ \approx 0.126 \text{ mol} \]
03
- Calculate the number of moles of hydrogen
Given the number of hydrogen atoms (\(3.8 \times 10^{23}\)), convert this quantity to moles. The molar amount of hydrogen atoms is determined by Avogadro’s number (\[6.022 \times 10^{23} \text{ atoms/mol}\]). \[ \text{Moles of H} = \frac{3.8 \times 10^{23} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mol}} \approx 0.631 \text{ mol} \]
04
- Calculate the mass of hydrogen
Using the molar mass of hydrogen (1.01 g/mol), calculate the mass of hydrogen from the moles found in Step 3. \[ \text{Mass of H} = 0.631 \text{ mol} \times 1.01 \text{ g/mol} \ \approx 0.637 \text{ g} \]
05
- Calculate carbon mass
Determine the mass of carbon in the compound by subtracting the masses of nitrogen and hydrogen from the total mass (10.0 g). \[ \text{Mass of C} = 10.0 \text{ g} - (1.77 \text{ g} + 0.637 \text{ g}) \approx 7.59 \text{ g} \]
06
- Calculate moles of carbon
Using the molar mass of carbon (12.01 g/mol), convert the mass of carbon found in Step 5 into moles. \[ \text{Moles of C} = \frac{7.59 \text{ g}}{12.01 \text{ g/mol}} \ \approx 0.632 \text{ mol} \]
07
- Determine simplest whole number ratio
Find the simplest ratio of moles of each element (carbon, hydrogen, and nitrogen) by dividing each by the smallest number of moles calculated: \[ \text{N:} \frac{0.126}{0.126} = 1 \]\[ \text{H:} \frac{0.631}{0.126} = 5 \]\[ \text{C:} \frac{0.632}{0.126} \ \approx 5 \]
08
- Write the empirical formula
Based on the ratio from Step 7 (which is close to 1:5:5 for C, H, and N respectively), the empirical formula of the compound is \ \[ \text{C}_{5}\text{H}_{5}\text{N} \].
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Moles Calculation
To calculate moles, you need to know the amount of substance you're working with in grams and the molar mass of the substance. The formula is pretty straightforward: \(\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\). For instance, when calculating the moles of nitrogen from 1.77 g N using its molar mass of 14.01 g/mol, you get approximately 0.126 mol N. This step is crucial for breaking down the composition of compounds and forms the foundation for further calculations.
Chemical Composition
Understanding chemical composition involves knowing the specific elements in a compound and their quantities. In the given exercise, the compound contains carbon, hydrogen, and nitrogen. With 10.0 g of the compound, we find that 17.7% is nitrogen and convert this to grams (1.77 g). Similarly, we identify the volume of hydrogen by the number of atoms and use Avogadro's number to convert it to moles. Finally, the remaining mass after accounting for nitrogen and hydrogen gives us the mass of carbon, which we then convert to moles. This method helps in understanding the makeup and proportions of different elements in a compound.
Molecular Formula
The empirical formula provides the simplest whole number ratio of the elements in a compound. However, the molecular formula gives the actual number of atoms of each element in a molecule. While the empirical formula from our exercise is \(\text{C}_{5}\text{H}_{5}\text{N}\), determining the molecular formula requires the actual molar mass of the compound, which can be a multiple of the empirical formula mass. For instance, if the molar mass is known to be 162 g/mol for this example, and the empirical formula mass is found to be 81 g/mol, the molecular formula will be \(\text{C}_{10}\text{H}_{10}\text{N}_{2}\).
Stoichiometry
Stoichiometry is the study of the quantitative relationships in chemical reactions. It involves using the balanced chemical equation to determine the proportions of reactants and products. In practice, stoichiometry includes calculations like those in our exercise, where we determine the amounts of each element in a compound through steps like moles calculation, converting masses, and simplifying ratios. This principle allows chemists to predict the quantities of products formed in a reaction, ensuring accurate and efficient chemical synthesis.
Avogadro's Number
Avogadro's number, \(6.022 \times 10^{23}\), is a fundamental constant used to count entities like atoms, ions, or molecules in a mole of any substance. It is vital for converting particles to moles and vice versa. For example, in the exercise, we used Avogadro's number to convert \(3.8 \times 10^{23}\) hydrogen atoms to 0.631 moles. This concept helps bridge the atomistic and macroscopic worlds in chemistry, enabling more practical computation and understanding of chemical quantities. Simply put, it lets us work with manageable quantities instead of dealing with astronomical numbers of atoms or molecules.
