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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:

<u>Part A: nicotine </u>

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 $C_5H_7N_1=C_5H_7N$

<u>Part B: caffeine</u>

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 $C_4H_5N_2O_1=C_4H_5N_2O$

6 0

3 years ago

Why is water said to be universal solvent?

Pachacha [2.7K]

Answer:

B

Explanation:

And, water is called the "universal solvent" because it dissolves more substances than any other liquid. This allows the water molecule to become attracted to many other different types of molecules.

8 0

2 years ago

Read 2 more answers

Please help I really appreciate it thank you ❤️

Feliz [49]

II. sulfur (S) and carbon (C)

and

III. fluorine (F) and oxygen (O)

will form covalent bonds, so the answer will be:

e. II and III

Explanation:

To know is what type of bond is formed between atoms we need to look at the electronegativity difference between the atoms.

If the electronegativity difference is less than 0.4 there is a nonpolar covalent bond.

If the electronegativity difference is between 0.4 and 1.8 there is a polar covalent bond. (if is a metal involved we consider the bond to be ionic)

If the electronegativity difference is greater then 1.8 there is an ionic bond.

We have the following cases:

I. lithium (Li) and sulfur (S)

electronegativity difference = 2.5 (S) - 1 (Li) = 1.5 but because there is a metal involved the bond will be ionic

II. sulfur (S) and carbon (C)

electronegativity difference = 2.5 (S) - 2.5 (C) = 0 so the bond will be nonpolar covalent

III. fluorine (F) and oxygen (O)

electronegativity difference = 4 (F) - 3.5 (O) = 0.5 so the bond will be polar covalent bond.

Learn more about:

covalent and ionic bonds

brainly.com/question/1802971

#learnwithBrainly

3 0

3 years ago

A sample of iron metal is placed in a graduated cylinder. it is noted that 10.4 ml of water is displaced by the iron. the iron i

Pavel [41]

<h3>Answer:</h3>

162.43 g of FeCl₂

<h3>Explanation:</h3>

Step 1: Calculate mass of Fe;

As,

Density = Mass ÷ Volume

Or,

Mass = Density × Volume

Where Volume is the volume of water displaced = 10.4 mL

Putting values,

Mass = 7.86 g.mL⁻¹ × 10.4 mL

Mass = 81.744 g of Fe

Step 2: Calculate amount of FeCl₂;

The balance chemical equation is as follow,

Fe + 2 HCl → FeCl₂ + H₂ ↑

According to this equation,

55.85 g (1 mol) Fe produced = 110.98 g (1 mol) of FeCl₂

So,

81.744 g Fe will produce = X g of FeCl₂

Solving for X,

X = (81.744 g × 110.98 g) ÷ 55.85 g

X = 162.43 g of FeCl₂

7 0

3 years ago

1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

andriy [413]

Answer:

A. $m_{NH_3}^{theo} =1.50gNH_3$

B. $Y=82.2\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

$n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\ n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3$

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

$m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3$

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

$Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%$

Best regards!

5 0

3 years ago