4Fe +302 + 2Fe2O3 How many grams of Fe2O3 will be produced from 10.0 g of Fe and 7.00 g of O2?

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

Part A: nicotine

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

Part B: caffeine

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

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Atom A has 32 protons and 35 neutrons. atom B has 32 protons and 33 neutrons. are they isotopes?

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Which SI unit indicates the mass on an average-sized box of breakfast cereal?

marin [14]

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Answer = (C.) g

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3 years ago

HELP AN ILL PUT U IN BRAINLIEST - THANKYOU

asambeis [7]

Answer:

D

Explanation:

The amount increased to a certain point then decreased

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2 years ago

Read 2 more answers

You arrive at a crime scene and are told the body of the victim is at the bottom of a pool that is 10 m deep. You walk over to w

monitta

Answer:

The vinegar is not enough to neutralize the pool.

Explanation:

The [OH⁻] in the pool is 1.0x10⁻¹mol / L. To know how many moles of OH⁻ are in the solution, you must calculate volume of the pool thus:

V(pool) = πr²h

Where radius is d/2 = 12m/2 = 6m

and h is deep of the pool = 10m

V(pool) = π(6m)²*10

V(pool) = 1131m³

As 1m³ = 1000L:

1131000L in the pool. And moles of OH⁻ are:

1.0x10⁻¹mol / L ₓ 1131000L = 131100 moles of OH⁻ are in the pool

The neutralization of OH⁻ with H⁺ is:

OH⁻ + H⁺ → H₂O

To neutralize the pool you must add 131100 moles of H⁺.

The H⁺ concentration in a vinegar pH = 2 is:

pH = -log [H⁺]

2 = -log [H⁺]

1x10⁻²M = [H⁺]

4L are just 4x10⁻² moles of [H⁺]. That means:

<h3>The vinegar is not enough to neutralize the pool.</h3>

5 0

3 years ago