## Chemistry 8 Lesson 21: Calculate according to the chemical formula

## 1. Summary of theory

### 1.1. Determine the percentage composition of the elements in a compound:

**a. What are the steps to determine the elemental composition of a compound? **

**– Method 1: **

Consider the chemical formula: AxByCz

\(\% A = \frac{{x. {M_A}}}{{{M_{hc}}}}.100\% \)

\(\% B = \frac{{y. {M_B}}}{{{M_{hc}}}}.100\% \)

C% = 100% – (%A + %B)

**– Method 2: **

Find the molar mass of the compound

Find the number of moles of each element in 1 mole of the compound and convert to mass.

Find the percentage composition of elements in a compound.

**b. Eg**

**Question:** A fertilizer with the formula KNO_{3} , determine the percent composition (by mass) of the elements.

**Tutorial:**

*** Method 1: Consider the chemical formula KNO**_{3}

Applying the formula we have:

\({\% N = \frac{{{M_N}}}{{{M_{hh}}}} = \frac{{14}}{{101}}.100 = 13.8\% }\)

\({\% K = \frac{{{M_K}}}{{{M_{hh}}}} = \frac{{39}}{{101}}.100 = 36.8.\% }\ )

%O = 100% – (13.8 + 36.8) = 47.8%

*** Method 2:**

**– Step 1: **Determine the molar mass of the compound.

MKNO_{3} = 39 + 14 + 16.3 = 101

**– Step 2:** Determine the number of moles of each element in 1 mole of the compound.

In 1mol KNO_{3 }Yes:

1 mole of K atoms ⇒ Potassium atomic mass is: 1 . 39 = 39

1 mole of N atom ⇒ Nitrogen atomic mass is: 1. 14 = 14

3 moles of O atoms ⇒ The mass of O atoms is: 16 . 3 = 48

**– Step 3:** Calculate the % composition of each element

\({\% K = \frac{{39}}{{101}}.100 = 36.8.\% }\)

\({\% N = \frac{{14}}{{101}}.100 = 13.8\% }\)

\({\% O = \frac{{48}}{{101}}.100 = 47.8\% }\)

### 1.2. Knowing the composition of the elements, determine the chemical formula of the compound

**a. Steps to determine the chemical formula of a compound**

**– Step 1:** Find the mass of each element present in 1 mole of the compound.

**– Step 2:** Find the number of moles of each element in 1 mole of the compound.

**– Step 3:** Determine the chemical formula of the compound.

**b. Eg**

**Question:** A compound has an elemental composition by mass of: 40% Cu; 20%S and 40%O. Determine the chemical formula of that substance. The compound has a molar mass of 160g/mol .

**Tutorial:**

Find the mass of each element present in 1 mole of the compound.

\(\begin{array}{*{20}{l}}

{{m_{Cu}} = \frac{{40}}{{100}}.160 = 64g}\\

{{m_S} = \frac{{20}}{{100}}.160 = 32g}\\

{{m_O} = \frac{{40}}{{100}}.160 = 64g}

\end{array}\)

Find the number of moles of each element present in 1 mole of the compound.

\(\begin{array}{*{20}{l}}

{{n_{Cu}} = \frac{{{m_{Cu}}}}{{{M_{Cu}}}} = \frac{{64}}{{64}} = 1mol}\\

{{n_S} = \frac{{{m_S}}}{{{M_S}}}{\rm{ = }}\frac{{32}}{{32}}{\rm{ = }}1mol}\ \

{{n_O} = \frac{{{m_O}}}{{{M_O}}}{\rm{ = }}\frac{{64}}{{16}}{\rm{ = }}4mol.}

\end{array}\)

The compound has a molar mass of 160g/mol and the number of moles of Cu, S, and O are as above, so the chemical formula of the compound is CuSO_{4}

## 2. Illustrated exercise

### 2.1. Form 1: When the mass % composition of elements and molecular mass is known

Determine the chemical composition of an oxide given that the molecular mass of the oxide is 160 and the percentage composition by mass of the element iron is 70%.

__Solution guide__

We call the formula of the oxide we are looking for FexOy .

– To calculate the x, y indices, we set up the mass ratios of the elements and compounds:

\(\frac{{x.56}}{{160}} = \frac{{70}}{{100}}\) and \(\frac{{y.16}}{{160}} = \ frac{{30}}{{100}}\)

and Infer: \(x = \frac{{160.70}}{{56.100}} = 2;y = \frac{{160.30}}{{16,100}} = 3\) ⇒ The formula for the oxide is: Fe2O3

### 2.2. Form 2: When mass ratios of elements and molecular masses are known

Determine the formula for an oxide of nitrogen, given that the molecular mass is 46 and the mass ratio mN : mO = 3.5 : 8

__Solution guide__

We call the chemical formula to find the form NxOy

– We have: x.14 + y.16 = 46 (1)

– Set up mass ratio: mN : mO = x . 14 : y . 16 = 3.5 : 8

Draw the ratio: x : y = (3.5/14):(8/16) = 0.25 : 0.5 = 1:2

Deduce: 2x = y, substitute (1) and solve to get: x = 1 and y = 2

⇒ The chemical formula of oxide is NO_{2}

### 2.3. Form 3: In the problem, it is possible to give the data to find the molecular mass

**Lesson 1: **Knowing the acid HxSyOz has %S = 32.65% and y = 1. Calculate the molar mass of the acid.

__Solution guide__

We have

\(\begin{array}{*{20}{l}}

{\% S = \frac{{y. {M_S}}}{{{M_{{H_x}{S_y}{O_z}}}}}.100\% }\\

{ \to {M_{{H_x}{S_y}{O_z}}} = \frac{{32.100}}{{32,65}} = 98}

\end{array}\)

⇒ So the molecular mass of acid HxSyOz is 98

**Lesson 2: **We know that 1 liter of acetylene gas (light earth gas) weighs 1.16g. Calculate the molar mass of acetylene gas

__Solution guide__

We have: V = n.22.4 n = V/22.4

on the other hand, we have: M = m/n

According to the problem we have: m = 1.16g, V = 1 liter

So the molar mass of acetylene gas is equal to M = 1.16.22,4 = 26 (g)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Calculate the mass of Fe in 92.8 g Fe_{3}O_{4}

**Verse 2:** Calculate m_{Al}_{2}O_{3} The number of moles of Al present in the compound is 0.6 .

**Question 3:** Ratio of moles of elements present in C_{3}H_{6}O_{2}

**Verse 4:** Find the chemical formula knowing that substance A has 80% Cu atoms and 20% oxygen atoms, knowing d_{A/HY}_{2} = 40?

**Question 5:** Calculate the mass percent of Mg present in 1 mole of MgO?

### 3.2. Multiple choice exercises

**Question 1:** For_{2}H_{5}OH. The number of H atoms in the compound

A. 1

B. 5

C. 3

D. 6

**Verse 2:** Calculate %m_{KY} present in the K molecule_{2}HAVE_{3}

A. 56, 502%

B. 56.52%

C. 56.3%

D. 56.56%

**Question 3:** Know which compounds have_{A/HY}_{2} = 22. Identify the compound known to have only 1 Oxygen atom

A. NO

B. CO

CN_{2}O

D. CO_{2}

**Verse 4:** Calculate %mC known in 1 mole of NaHCO_{3} there are 1 mol Na, 1 mol C and 3 mol O, 1 mol H

A. 14.28%

B. 14.2%

C. 14.284%

D. 14.285%

**Question 5:** Mass percent composition of oxygen in Fe_{2}O_{3}

A. 35%

B. 40%

C. 30%

D. 45%

## 4. Conclusion

After the lesson need:

- Calculate the percent composition by mass of the elements present in a compound when the chemical formula of that compound is known.

.

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