What is the concentration of sugar ($C_{12}H_{22}O_{11}$) in $mol\ L^{-1}$ if its 20 g are dissolved in enough water to make a final volume up to 2 L?

What is the concentration of sugar (C₁₂H₂₂O₁₁) in molL⁻¹ if its 20 g are dissolved in enough water to make a f

Exercises · 1.11

C_{12}H_{22}O_{11}

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Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.

Atoms and molecules are unimaginably tiny. A single water molecule ( $H_2O$ ) weighs about $3 \times 10^{-23}$ grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, $6.022 \times 10^{23}$ molecules, which is called one mole.

Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have $6.022 \times 10^{23}$ molecules of this compound, how much would they weigh on a lab balance?"

Note

The number $6.022 \times 10^{23}$ is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.

The Precise Definition

Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.

For example:

Water ( $H_2O$ ): 2 hydrogen atoms + 1 oxygen atom
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molecular mass of $H_2O$ = $2(1.008) + 16.00 = 18.016$ g/mol

\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass of that element})

How to Calculate It — Step by Step

Let's take glucose, $C_6H_{12}O_6$ , as a worked example.

Step 1: Identify each element and its count

Carbon (C): 6 atoms
Hydrogen (H): 12 atoms
Oxygen (O): 6 atoms

Step 2: Look up atomic masses (from the periodic table)

C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol

Step 3: Multiply and add

\text{Molecular mass} = 6(12.01) + 12(1.008) + 6(16.00)

= 72.06 + 12.096 + 96.00

= 180.156 \text{ g/mol}

Tip

Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.

Why This Matters

Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:

Convert grams to moles: $\text{moles} = \frac{\text{mass in grams}}{\text{molecular mass}}$
Convert moles to grams: $\text{mass} = \text{moles} \times \text{molecular mass}$
Determine the number of molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

Watch out

Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like $CO_2$ = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).

Common Exam Pitfalls

Forgetting to multiply by the subscript. In $H_2SO_4$ , there are 2 hydrogens, not 1. …

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.

1. The Core Question: What is a Mole?

A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains $6.022 \times 10^{23}$ particles (Avogadro's number, $N_A$ ).

Why this number?

It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.

Example: 1 atom of carbon-12 has mass 12 amu.
1 mole of carbon-12 has mass 12 grams.

This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).

2. The Fundamental Relationship

The key formula is:

n = \frac{m}{M}

Where:

$n$ = number of moles
$m$ = mass of substance (in grams)
$M$ = molar mass (in g/mol)

Why does this work?

Think of it as a conversion factor:

If 1 mole of a substance weighs $M$ grams, then $m$ grams contains $\frac{m}{M}$ moles.

Derivation logic:

Molar mass $M$ tells you: "1 mol = $M$ g"
So the conversion factor is $\frac{1 \text{ mol}}{M \text{ g}}$
Multiply mass $m$ by this factor: $m \times \frac{1}{M} = \frac{m}{M}$ moles

3. Connecting to Number of Particles

n = \frac{N}{N_A}

Where:

$N$ = number of particles (atoms, molecules, ions)
$N_A = 6.022 \times 10^{23}$ particles/mol

Why?

1 mole = $N_A$ particles
So $N$ particles = $\frac{N}{N_A}$ moles

Combine both formulas:

\frac{m}{M} = \frac{N}{N_A}

This single equation ties mass, molar mass, number of particles, and Avogadro's number together.

4. The Gas Volume Connection (for gases at STP)

For gases only:

n = \frac{V}{22.4 \text{ L/mol}}

Why 22.4 L?

From the ideal gas law: $PV = nRT$

At STP (Standard Temperature and Pressure: 0°C, 1 atm):

$P = 1$ atm
$T = 273.15$ K
$R = 0.0821$ L·atm/(mol·K)

For $n = 1$ mole:

V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273.15}{1} \approx 22.4 \text{ L}

So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.

5. The Stoichiometry Chain: From One Substance to Another

In a balanced chemical equation like:

aA + bB \rightarrow cC + dD

The coefficients tell you the mole ratio:

\frac{\text{moles of A}}{\text{moles of B}} = \frac{a}{b}

Why this works: …

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To find the concentration of sugar, first calculate its molar mass, then use it to convert the given mass of sugar into moles. Finally, divide the moles by the total volume of the solution to get the molarity. The concentration is $\boxed{0.029 \text{ mol L}^{-1}}$ .

When we talk about the "concentration" of a solution in chemistry, we often mean its molarity. Molarity is a measure of how many moles of a solute are dissolved in one liter of solution. It's a very common way to express concentration because it directly relates to the number of particles (molecules or ions) present, which is what drives chemical reactions.

To calculate molarity, we need two key pieces of information:

The number of moles of the solute.
The total volume of the solution in liters.

The problem gives us the mass of sugar and the final volume of the solution. Our task is to convert the given mass of sugar into moles, and then use the provided volume to find the molarity.

Here's how we approach this problem step-by-step:

Identify the given information.

We are given:
- Mass of sugar ( $C_{12}H_{22}O_{11}$ ) = 20 g
- Final volume of solution = 2 L
Determine the molecular formula and calculate the molar mass of sugar.

The molecular formula for sugar (sucrose) is $C_{12}H_{22}O_{11}$ . To convert the mass of sugar into moles, we first need to calculate its molar mass. The molar mass is the sum of the atomic masses of all atoms in one molecule. We'll use the standard atomic masses:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol
- Oxygen (O): 15.999 g/mol
The molar mass of $C_{12}H_{22}O_{11}$ is calculated as follows:

\text{Molar mass} = (12 \times \text{Atomic mass of C}) + (22 \times \text{Atomic mass of H}) + (11 \times \text{Atomic mass of O})

\text{Molar mass} = (12 \times 12.01 \text{ g/mol}) + (22 \times 1.008 \text{ g/mol}) + (11 \times 15.999 \text{ g/mol})

\text{Molar mass} = 144.12 \text{ g/mol} + 22.176 \text{ g/mol} + 175.989 \text{ g/mol}

\text{Molar mass} = 342.285 \text{ g/mol}

Calculate the number of moles of sugar. …

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Common Mistakes in Molecular Mass Calculation & Concentration Problems

Students often lose marks on this simple problem due to avoidable errors. Here are the most frequent mistakes and how to fix each.

✗ Mistake 1: Incorrect Molecular Mass Calculation

What students do wrong:

They mis-count the number of atoms or use wrong atomic masses. Common errors:

Using $C = 12$ , $H = 1$ , $O = 16$ but forgetting to multiply by the number of atoms
Adding atomic masses without multiplying (e.g., $12 + 1 + 16$ instead of $12 \times 12 + 22 \times 1 + 11 \times 16$ )

How to avoid:

Write the formula clearly and calculate step-by-step:

C_{12}H_{22}O_{11}

Carbon: $12 \times 12 = 144$
Hydrogen: $22 \times 1 = 22$
Oxygen: $11 \times 16 = 176$

Total molar mass = $144 + 22 + 176 = \mathbf{342\ g\ mol^{-1}}$

Tip: Always double-check the subscript numbers against the formula.

✗ Mistake 2: Confusing Mass with Moles

What students do wrong:

They plug 20 g directly into the molarity formula without converting to moles first.

How to avoid:

Remember: Molarity ( $M$ ) = moles of solute / volume in litres

First convert mass to moles:

\text{Moles of sugar} = \frac{\text{mass}}{\text{molar mass}} = \frac{20}{342} \approx 0.0585\ \text{mol}

✗ Mistake 3: Volume Unit Errors

What students do wrong:

Using volume in mL without converting to L
Forgetting that the final volume is 2 L (not 1 L)

How to avoid:

Always check the unit of volume given. Here it's already in litres — 2 L.

If volume were in mL, divide by 1000.

✗ Mistake 4: Forgetting the "Final Volume" Concept

What students do wrong:

They think "2 L of water is added" instead of "final volume is 2 L". This changes the calculation.

How to avoid: …

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Q.What is the concentration of sugar ( $C_{12}H_{22}O_{11}$ ) in $mol\ L^{-1}$ if its 20 g are dissolved in enough water to make a final volume up to 2 L?

Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

The Precise Definition

How to Calculate It — Step by Step

Why This Matters

Common Exam Pitfalls

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

1. The Core Question: What is a Mole?

2. The Fundamental Relationship

Why does this work?

3. Connecting to Number of Particles

Why?

4. The Gas Volume Connection (for gases at STP)

Why 22.4 L?

5. The Stoichiometry Chain: From One Substance to Another

Why this works: …

Method: Molarity Formula Method

Steps

Common Mistakes in Molecular Mass Calculation & Concentration Problems

✗ Mistake 1: Incorrect Molecular Mass Calculation

✗ Mistake 2: Confusing Mass with Moles

✗ Mistake 3: Volume Unit Errors

✗ Mistake 4: Forgetting the "Final Volume" Concept

Q.What is the concentration of sugar (C12H22O11C_{12}H_{22}O_{11}C12​H22​O11​) in mol L−1mol\ L^{-1}mol L−1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?

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Q.What is the concentration of sugar ( $C_{12}H_{22}O_{11}$ ) in $mol\ L^{-1}$ if its 20 g are dissolved in enough water to make a final volume up to 2 L?

Unlock everything free for 14 days