Math / Algebra

Question3. Nicotine $\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{~N}_{z}$ is a harmful compound. It affects the respiratory system and nervous system with many problems. It causes many diseases. The analysis of nicotine shows that the $\%$ by mass of carbon is $74.07 \%$ where 32.4 g nicotine contains 2.8 g hydrogen. Given: Molar masses (g. $\mathrm{mol}^{-1}$ ): N-14, C-12, H-1. Avogadro's number: $\mathrm{N}_{\mathrm{A}}=\mathbf{6} \times 10^{23}$ particles / mol. 3.1- Determine the $\%$ by mass of hydrogen and nitrogen in nicotine. 3.2- Determine the empirical formula of nicotine. 3.3- Show that the molecular formula of nicotine is $\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}$ if $\mathrm{M}_{\text {Niootine }}=162 \mathrm{~g} \cdot \mathrm{~mol}^{1}$ . 3.4 A cigarette contains 16.2 g of nicotine. Determine the number of carbon and hydrogen atoms present in the above mass.

Studdy Solution

STEP 1

What is this asking? We're figuring out the makeup of nicotine – how much of each element it contains, its simplest formula, its actual formula, and how many atoms are in a cigarette's worth. Watch out! Don't mix up empirical and molecular formulas!
The empirical formula is the simplest ratio, while the molecular formula tells you the actual number of atoms in a molecule.

STEP 2

1. Find the Percentage by Mass
2. Determine the Empirical Formula
3. Verify the Molecular Formula
4. Calculate the Number of Atoms

STEP 3

We know that 32.4 g of nicotine contains 2.8 g of hydrogen.
So, the percentage of hydrogen by mass is: $\frac{2.8 \text{ g H}}{32.4 \text{ g Nicotine}} \cdot 100\% \approx 8.64\% \text{ H}$ So, hydrogen makes up roughly 8.64% of nicotine by mass.

STEP 4

Since we know the percentages of carbon and hydrogen add up to 100% with nitrogen, we can find the percentage of nitrogen: $100\% - 74.07\% \text{ C} - 8.64\% \text{ H} \approx 17.29\% \text{ N}$ Nitrogen makes up approximately 17.29% of nicotine.

STEP 5

Let's imagine we have a 100 g sample of nicotine.
This makes the percentages directly translate to grams.
So, we have approximately 74.07 g of carbon, 8.64 g of hydrogen, and 17.29 g of nitrogen.

STEP 6

We'll use the molar masses to convert these grams into moles. Carbon: $\frac{74.07 \text{ g C}}{12 \text{ g/mol}} \approx 6.17 \text{ mol C}$ Hydrogen: $\frac{8.64 \text{ g H}}{1 \text{ g/mol}} \approx 8.64 \text{ mol H}$ Nitrogen: $\frac{17.29 \text{ g N}}{14 \text{ g/mol}} \approx 1.24 \text{ mol N}$

STEP 7

Divide each mole value by the smallest one (1.24 mol N) to get the simplest ratio: Carbon: $\frac{6.17}{1.24} \approx 5$ Hydrogen: $\frac{8.64}{1.24} \approx 7$ Nitrogen: $\frac{1.24}{1.24} = 1$ This gives us an empirical formula of C₅H₇N.

STEP 8

The empirical formula mass of C₅H₇N is: $(5 \cdot 12) + (7 \cdot 1) + (1 \cdot 14) = 60 + 7 + 14 = 81 \text{ g/mol}$

STEP 9

Divide the given molecular mass of nicotine (162 g/mol) by the empirical formula mass (81 g/mol): $\frac{162}{81} = 2$ This means the molecular formula is twice the empirical formula.

STEP 10

Multiply the subscripts in the empirical formula by 2: $\text{C}_{5 \cdot 2}\text{H}_{7 \cdot 2}\text{N}_{1 \cdot 2} = \text{C}_{10}\text{H}_{14}\text{N}_{2}$ This confirms the molecular formula as C₁₀H₁₄N₂.

STEP 11

A cigarette contains 16.2 g of nicotine.
Convert this to moles: $\frac{16.2 \text{ g}}{162 \text{ g/mol}} = 0.1 \text{ mol}$

STEP 12

Multiply the moles of nicotine by Avogadro's number: $0.1 \text{ mol} \cdot 6 \times 10^{23} \text{ molecules/mol} = 6 \times 10^{22} \text{ molecules}$

STEP 13

Each molecule of nicotine has 10 carbon atoms and 14 hydrogen atoms.
So, in $6 \times 10^{22}$ molecules: Carbon atoms: $10 \cdot 6 \times 10^{22} = 6 \times 10^{23}$ atoms Hydrogen atoms: $14 \cdot 6 \times 10^{22} = 8.4 \times 10^{23}$ atoms

STEP 14

The mass percentages are approximately 8.64% hydrogen and 17.29% nitrogen.
The empirical formula is C₅H₇N.
The molecular formula is confirmed as C₁₀H₁₄N₂.
A cigarette containing 16.2 g of nicotine has $6 \times 10^{23}$ carbon atoms and $8.4 \times 10^{23}$ hydrogen atoms.

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Studdy Solution

STEP 1

STEP 2

1. Find the Percentage by Mass
2. Determine the Empirical Formula
3. Verify the Molecular Formula
4. Calculate the Number of Atoms

STEP 3

We know that 32.4 g of nicotine contains 2.8 g of hydrogen.
So, the percentage of hydrogen by mass is: $\frac{2.8 \text{ g H}}{32.4 \text{ g Nicotine}} \cdot 100\% \approx 8.64\% \text{ H}$ So, hydrogen makes up roughly 8.64% of nicotine by mass.

STEP 4

Since we know the percentages of carbon and hydrogen add up to 100% with nitrogen, we can find the percentage of nitrogen: $100\% - 74.07\% \text{ C} - 8.64\% \text{ H} \approx 17.29\% \text{ N}$ Nitrogen makes up approximately 17.29% of nicotine.

STEP 5

Let's imagine we have a 100 g sample of nicotine.
This makes the percentages directly translate to grams.
So, we have approximately 74.07 g of carbon, 8.64 g of hydrogen, and 17.29 g of nitrogen.

STEP 6

STEP 7

Divide each mole value by the smallest one (1.24 mol N) to get the simplest ratio: Carbon: $\frac{6.17}{1.24} \approx 5$ Hydrogen: $\frac{8.64}{1.24} \approx 7$ Nitrogen: $\frac{1.24}{1.24} = 1$ This gives us an empirical formula of C₅H₇N.

STEP 8

The empirical formula mass of C₅H₇N is: $(5 \cdot 12) + (7 \cdot 1) + (1 \cdot 14) = 60 + 7 + 14 = 81 \text{ g/mol}$

STEP 9

Divide the given molecular mass of nicotine (162 g/mol) by the empirical formula mass (81 g/mol): $\frac{162}{81} = 2$ This means the molecular formula is twice the empirical formula.

STEP 10

Multiply the subscripts in the empirical formula by 2: $\text{C}_{5 \cdot 2}\text{H}_{7 \cdot 2}\text{N}_{1 \cdot 2} = \text{C}_{10}\text{H}_{14}\text{N}_{2}$ This confirms the molecular formula as C₁₀H₁₄N₂.

STEP 11

A cigarette contains 16.2 g of nicotine.
Convert this to moles: $\frac{16.2 \text{ g}}{162 \text{ g/mol}} = 0.1 \text{ mol}$

STEP 12

Multiply the moles of nicotine by Avogadro's number: $0.1 \text{ mol} \cdot 6 \times 10^{23} \text{ molecules/mol} = 6 \times 10^{22} \text{ molecules}$

STEP 13

Each molecule of nicotine has 10 carbon atoms and 14 hydrogen atoms.
So, in $6 \times 10^{22}$ molecules: Carbon atoms: $10 \cdot 6 \times 10^{22} = 6 \times 10^{23}$ atoms Hydrogen atoms: $14 \cdot 6 \times 10^{22} = 8.4 \times 10^{23}$ atoms

STEP 14

The mass percentages are approximately 8.64% hydrogen and 17.29% nitrogen.
The empirical formula is C₅H₇N.
The molecular formula is confirmed as C₁₀H₁₄N₂.
A cigarette containing 16.2 g of nicotine has $6 \times 10^{23}$ carbon atoms and $8.4 \times 10^{23}$ hydrogen atoms.

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Studdy Solution

STEP 1

STEP 2

1. Find the Percentage by Mass2. Determine the Empirical Formula3. Verify the Molecular Formula4. Calculate the Number of Atoms

STEP 3

STEP 4

STEP 5

Let's imagine we have a **100 g** sample of nicotine.This makes the percentages directly translate to grams.So, we have approximately 74.07 g of carbon, 8.64 g of hydrogen, and 17.29 g of nitrogen.

STEP 6

STEP 7

STEP 8

The empirical formula mass of C5H7N is: (5⋅12)+(7⋅1)+(1⋅14)=60+7+14=81 g/mol (5 \cdot 12) + (7 \cdot 1) + (1 \cdot 14) = 60 + 7 + 14 = 81 \text{ g/mol} (5⋅12)+(7⋅1)+(1⋅14)=60+7+14=81 g/mol

STEP 9

Divide the given molecular mass of nicotine (162 g/mol) by the empirical formula mass (81 g/mol): 16281=2 \frac{162}{81} = 2 81162​=2 This means the molecular formula is twice the empirical formula.

STEP 10

Multiply the subscripts in the empirical formula by 2: C5⋅2H7⋅2N1⋅2=C10H14N2 \text{C}_{5 \cdot 2}\text{H}_{7 \cdot 2}\text{N}_{1 \cdot 2} = \text{C}_{10}\text{H}_{14}\text{N}_{2} C5⋅2​H7⋅2​N1⋅2​=C10​H14​N2​ This confirms the molecular formula as **C10H14N2**.

STEP 11

A cigarette contains 16.2 g of nicotine.Convert this to moles: 16.2 g162 g/mol=0.1 mol \frac{16.2 \text{ g}}{162 \text{ g/mol}} = 0.1 \text{ mol} 162 g/mol16.2 g​=0.1 mol

STEP 12

Multiply the moles of nicotine by Avogadro's number: 0.1 mol⋅6×1023 molecules/mol=6×1022 molecules 0.1 \text{ mol} \cdot 6 \times 10^{23} \text{ molecules/mol} = 6 \times 10^{22} \text{ molecules} 0.1 mol⋅6×1023 molecules/mol=6×1022 molecules

STEP 13

STEP 14

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1. Find the Percentage by Mass
2. Determine the Empirical Formula
3. Verify the Molecular Formula
4. Calculate the Number of Atoms

Let's imagine we have a 100 g sample of nicotine.
This makes the percentages directly translate to grams.
So, we have approximately 74.07 g of carbon, 8.64 g of hydrogen, and 17.29 g of nitrogen.

The empirical formula mass of C₅H₇N is: $(5 \cdot 12) + (7 \cdot 1) + (1 \cdot 14) = 60 + 7 + 14 = 81 \text{ g/mol}$

Divide the given molecular mass of nicotine (162 g/mol) by the empirical formula mass (81 g/mol): $\frac{162}{81} = 2$ This means the molecular formula is twice the empirical formula.

Multiply the subscripts in the empirical formula by 2: $\text{C}_{5 \cdot 2}\text{H}_{7 \cdot 2}\text{N}_{1 \cdot 2} = \text{C}_{10}\text{H}_{14}\text{N}_{2}$ This confirms the molecular formula as C₁₀H₁₄N₂.

A cigarette contains 16.2 g of nicotine.
Convert this to moles: $\frac{16.2 \text{ g}}{162 \text{ g/mol}} = 0.1 \text{ mol}$

Multiply the moles of nicotine by Avogadro's number: $0.1 \text{ mol} \cdot 6 \times 10^{23} \text{ molecules/mol} = 6 \times 10^{22} \text{ molecules}$