Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar

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CH_3COONa

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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 prepare a solution of a specific molarity, we first determine the required number of moles of solute using the given volume and desired molarity, then convert these moles into mass using the molar mass. The mass of sodium acetate needed is 15.38 g.

To calculate the mass of sodium acetate required, we need to understand the concept of molarity. Molarity is a measure of the concentration of a solute in a solution, specifically defined as the number of moles of solute per liter of solution. Our goal is to find the mass, but the given information (volume and molarity) directly leads us to the number of moles. Once we have the moles, we can use the molar mass to convert moles into mass.

Here's how we approach the calculation:

Identify the given information and the target quantity.

We are given:
- Volume of solution ( $V$ ) = $500\ mL$
- Molarity of solution ( $M$ ) = $0.375\ M$ (which means $0.375\ mol\ L^{-1}$ )
- Molar mass of sodium acetate ( $M_m$ ) = $82.0245\ g\ mol^{-1}$ We need to find the mass of sodium acetate ( $m$ ) required.
Convert the volume of the solution to liters.

Molarity is defined in moles per liter, so it is essential to express the volume in liters to maintain unit consistency in our calculations.

V = 500\ mL \times \frac{1\ L}{1000\ mL} = 0.500\ L

> [!WARNING]
> A common mistake is to use the volume in milliliters directly in the molarity formula. Always ensure your volume is in liters when working with molarity.

3. Calculate the number of moles of sodium acetate required.

The definition of molarity provides the relationship between moles, volume, and molarity:

> [!FORMULA]

> $M = \frac{\text{Number of moles of solute (n)}}{\text{Volume of solution in liters (V)}}$

We can rearrange this formula to solve for the number of moles ( $n$ ):

n = M \times V

Substitute the given molarity and the converted volume:

n = 0.375\ \frac{mol}{L} \times 0.500\ L

$$n = 0.1875\ mol$$ …

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Here are the most common mistakes students make when solving this exact problem, along with clear, exam-smart ways to avoid each.

1. Forgetting to Convert Volume to Litres

The Mistake

Students plug in 500 directly into the formula $M = \frac{n}{V}$ , treating volume in mL as if it were in litres. This gives a wildly wrong answer.

Why It’s Wrong

Molarity is defined as moles per litre. Volume must be in litres.

$500\ \text{mL} = 0.500\ \text{L}$ .

How to Avoid

Always write the unit next to every number.
Before using any formula, check: Is volume in litres? If not, convert immediately.
Quick trick: divide mL by 1000 → $500 \div 1000 = 0.5$ .

2. Using the Wrong Formula or Rearranging Incorrectly

The Mistake

Students sometimes use $n = M \times V$ but then forget to multiply by molar mass, or they mix up the formula for mass.

Why It’s Wrong

The correct chain is:

n = M \times V \quad \text{(moles = molarity × volume in L)}

\text{mass} = n \times \text{molar mass}

How to Avoid

Write the two-step process clearly:
1. Find moles: $n = 0.375\ \text{mol/L} \times 0.500\ \text{L} = 0.1875\ \text{mol}$
2. Find mass: $\text{mass} = 0.1875\ \text{mol} \times 82.0245\ \text{g/mol}$
Never skip writing the units — they act as a built-in check.

3. Rounding Too Early (Losing Precision)

The Mistake

Students round $0.375 \times 0.5 = 0.1875$ to $0.19$ or $0.2$ before multiplying by molar mass, then get a final answer like $15.6\ \text{g}$ instead of $15.3796\ \text{g}$ .

Why It’s Wrong

In competitive exams, small rounding errors can cost marks. The given molar mass has 6 significant figures — your answer should reflect that precision.

How to Avoid

Keep all intermediate values in your calculator.
Round only the final answer to the appropriate significant figures (here, 4 significant figures: $15.38\ \text{g}$ ).
If doing manual calculation, keep at least 4 decimal places until the end.

4. Confusing Molar Mass with Molecular Weight in Different Units

The Mistake

Some students treat molar mass as $82.0245\ \text{g}$ (without per mole) and think that’s the mass of one molecule, or they forget the unit $\text{g mol}^{-1}$ .

Why It’s Wrong

Molar mass is grams per mole. It’s a conversion factor, not a fixed mass.

How to Avoid …

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Q.Calculate the mass of sodium acetate ( $CH_3COONa$ ) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is $82.0245\ g\ mol^{-1}$ .

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

Concept (Why this works)

Steps

1. Forgetting to Convert Volume to Litres

2. Using the Wrong Formula or Rearranging Incorrectly

3. Rounding Too Early (Losing Precision)

4. Confusing Molar Mass with Molecular Weight in Different Units

Q.Calculate the mass of sodium acetate (CH3COONaCH_3COONaCH3​COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol−182.0245\ g\ mol^{-1}82.0245 g mol−1.

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Q.Calculate the mass of sodium acetate ( $CH_3COONa$ ) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is $82.0245\ g\ mol^{-1}$ .

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