Grade 10

Advanced

7 min

Lesson 5 of 12

Molecular Formulas and Molar Mass

Not Started

Moving from the simplest ratio to the actual chemical formula of a molecule using molar mass data.

Imagine finding a mysterious white powder at a crime scene. Chemical analysis tells you it's 40% carbon, but is it life-sustaining sugar or a deadly toxin? To know for sure, you must move beyond simple ratios and find the molecule's true identity.

Learning Objectives

1.Calculate the whole-number multiplier ( $n$ ) using the relationship between empirical mass and molar mass
2.Determine the actual molecular formula of a compound from experimental data
3.Analyze combustion data to reconstruct the identity of unknown hydrocarbons

The Whole-Number Multiplier

The empirical formula represents the simplest whole-number ratio of atoms in a compound. However, nature often builds molecules that are multiples of this ratio. The molecular formula is the actual number of atoms of each element in a molecule. To find it, we use a multiplier,

n

. This multiplier represents how many 'empirical units' fit into the actual molecule. Because you cannot have half an atom,

n

must always be a whole number. The relationship is expressed as:

Molecular\ Formula = n \times (Empirical\ Formula)

A compound has the empirical formula $CH_2$ and a molar mass of $42.0\ g/mol$ . Find its molecular formula.

1. Calculate the Empirical Formula Mass (EFM): $12.01\ (C) + 2 \times 1.01\ (H) = 14.03\ g/mol$ . 2. Use the formula $n = \frac{Molar\ Mass}{EFM}$ . 3. $n = \frac{42.0}{14.03} \approx 3$ . 4. Multiply the empirical formula by 3: $3 \times (CH_2) = C_3H_6$ .

Quick Check

If the empirical formula is $HO$ and the multiplier $n$ is 2, what is the molecular formula?

Multiply each subscript in

H_1O_1

by 2.

Answer

$H_2O_2$ (Hydrogen Peroxide)

The Bridge: Molar Mass

In a lab, we often determine the percentage composition first to find the empirical formula. But that only gives us the ratio. To find the identity, we need the molar mass, usually determined via mass spectrometry. The molar mass acts as a scale: if the empirical formula is a single brick, the molar mass tells us how heavy the entire building is. By dividing the total weight by the weight of one brick, we find exactly how many bricks ( $n$ ) were used to build the molecule.

A compound is $85.7\%$ Carbon and $14.3\%$ Hydrogen. Its molar mass is $28.0\ g/mol$ . Find the molecular formula.

1. Assume $100\ g$ : $85.7\ g\ C$ and $14.3\ g\ H$ . 2. Convert to moles: $C = \frac{85.7}{12.01} = 7.14\ mol$ ; $H = \frac{14.3}{1.01} = 14.16\ mol$ . 3. Find ratio: $\frac{14.16}{7.14} \approx 2$ . Empirical Formula = $CH_2$ . 4. EFM of $CH_2 = 14.03\ g/mol$ . 5. $n = \frac{28.0}{14.03} = 2$ . 6. Molecular Formula = $C_2H_4$ .

Quick Check

Can the molecular formula ever be the same as the empirical formula?

Think about molecules like

H_2O

CH_4

Answer

Yes, if the multiplier $n$ is equal to 1.

Advanced: Combustion Analysis

For unknown organic compounds, scientists use combustion analysis. By burning a sample in pure oxygen, all Carbon turns into $CO_2$ and all Hydrogen turns into $H_2O$ . By measuring the mass of these products, we can work backward to find the moles of $C$ and $H$ in the original sample. This is the gold standard for identifying unknown hydrocarbons in forensic and environmental chemistry.

Burning $0.44\ g$ of a hydrocarbon produced $1.32\ g$ of $CO_2$ and $0.72\ g$ of $H_2O$ . The molar mass is $44.1\ g/mol$ . Find the molecular formula.

1. Moles of $C$ from $CO_2$ : $\frac{1.32\ g}{44.01\ g/mol} = 0.03\ mol$ . 2. Moles of $H$ from $H_2O$ : $\frac{0.72\ g}{18.02\ g/mol} = 0.04\ mol\ H_2O$ . Since each $H_2O$ has $2\ H$ , moles of $H = 0.08\ mol$ . 3. Ratio $C:H = 0.03:0.08 = 3:8$ . Empirical Formula = $C_3H_8$ . 4. EFM of $C_3H_8 = (3 \times 12.01) + (8 \times 1.01) = 44.11\ g/mol$ . 5. $n = \frac{44.1}{44.11} = 1$ . Molecular Formula = $C_3H_8$ (Propane).

Key Takeaways

1The molecular formula is always a whole-number multiple of the empirical formula.
2The multiplier $n$ is calculated as $n = \frac{Molar\ Mass}{Empirical\ Formula\ Mass}$ .
3Molar mass data is essential to distinguish between different compounds that share the same empirical ratio.
4Combustion analysis allows us to determine formulas by measuring the mass of $CO_2$ and $H_2O$ produced.

Quick Check

Multiple Choice

A compound has an empirical formula of $CH$ and a molar mass of $78\ g/mol$ . What is its molecular formula?

Multiple Choice

If the molar mass of a compound is $180\ g/mol$ and its empirical formula mass is $30\ g/mol$ , what is the value of $n$ ?

True/False

In combustion analysis, the number of moles of Hydrogen atoms in the sample is equal to the number of moles of $H_2O$ produced.

What's Next?

Review Tomorrow

In 24 hours, try to explain to someone why $CH_2O$ could be either formaldehyde or glucose, and what specific piece of data you would need to tell them apart.

Practice Activity

Find the molecular formula for a compound with $40\%$ C, $6.7\%$ H, and $53.3\%$ O with a molar mass of $60\ g/mol$ .

Browse

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Molecular Formulas and Molar Mass

Learning Objectives

The Whole-Number Multiplier

The Bridge: Molar Mass

Advanced: Combustion Analysis

Key Takeaways

Quick Check

What's Next?

Molecular Formulas and Molar Mass

Learning Objectives

The Whole-Number Multiplier

The Bridge: Molar Mass

Advanced: Combustion Analysis

Key Takeaways

Quick Check

What's Next?