Calcium carbonate reacts with aqueous HCl to give $CaCl_2$ and $CO_2$ according to the reaction, $CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + H_2O (l)$. What mass of $CaCO_3$ is required to react completely with 25 mL of 0.75 M HCl?

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃ (s) + 2 HCl (

Exercises · 1.35

CaCl_2

(g) + H_2O (l)

. What mass of

CaCO_3$ is required to react completely with 25 mL of 0.75 M HCl?

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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 mass of $CaCO_3$ needed, we first calculate the moles of $HCl$ from its given volume and molarity, then use the stoichiometric ratio from the balanced equation to find the moles of $CaCO_3$ , and finally convert these moles to mass using the molar mass of $CaCO_3$ . The required mass of $CaCO_3$ is $\boxed{0.94 \text{ g}}$ .

When chemicals react, they do so in specific proportions. These proportions are not based on mass directly, but on the number of particles (atoms, molecules, or ions) involved. The concept of the 'mole' allows us to count these particles and relate them to a measurable mass. The balanced chemical equation provides the exact ratio in which moles of reactants combine and products form. This is the fundamental principle of stoichiometry.

In this problem, we are given the volume and concentration of $HCl$ , which allows us to determine the number of moles of $HCl$ available. Once we know the moles of $HCl$ , the balanced chemical equation will tell us how many moles of $CaCO_3$ are required to react completely with it. Finally, we can convert these moles of $CaCO_3$ into a mass using its molar mass.

Here is the step-by-step solution:

Understand the balanced chemical equation. The given balanced chemical equation is:

CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + H_2O (l)

This equation tells us that $1$ mole of solid calcium carbonate ($CaCO_3$) reacts completely with $2$ moles of aqueous hydrochloric acid ($HCl$). This $1:2$ mole ratio is crucial for our calculations.

2. Calculate the moles of $HCl$ available.

We are given the volume and molarity of the $HCl$ solution.

* Volume of $HCl = 25 \text{ mL}$

* Molarity of $HCl = 0.75 \text{ M}$ (which means $0.75 \text{ mol/L}$ )

First, convert the volume from milliliters to liters:

\text{Volume of } HCl = 25 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.025 \text{ L}

Now, use the definition of molarity to find the moles of $HCl$:
> [!FORMULA]
> $\text{Moles} = \text{Molarity} \times \text{Volume (in L)}$

\text{Moles of } HCl = 0.75 \frac{\text{mol}}{\text{L}} \times 0.025 \text{ L} = 0.01875 \text{ mol}

Determine the moles of $CaCO_3$ required. From the balanced equation, we know that $1$ mole of $CaCO_3$ reacts with $2$ moles of $HCl$ . We can use this stoichiometric ratio to find the moles of $CaCO_3$ needed to react with $0.01875$ moles of $HCl$ .

\text{Moles of } CaCO_3 = \text{Moles of } HCl \times \frac{1 \text{ mol } CaCO_3}{2 \text{ mol } HCl}

\text{Moles of } CaCO_3 = 0.01875 \text{ mol } HCl \times \frac{1}{2} = 0.009375 \text{ mol } CaCO_3

Calculate the molar mass of $CaCO_3$ . …

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Common Mistakes in Molecular Mass Calculation Problems (HCl & CaCO₃ Reaction)

Here are the most frequent errors students make in this type of problem, along with how to avoid each.

1. Forgetting the Mole Ratio from the Balanced Equation

The Mistake:

Students often assume 1 mole of HCl reacts with 1 mole of CaCO₃, leading to an answer that is double the correct value.

Why It Happens:

They see "HCl" and "CaCO₃" and skip checking the coefficients.

How to Avoid:

Always write the balanced equation and circle the coefficients before starting any calculation.

CaCO_3 (s) + \mathbf{2} HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + H_2O (l)

The key ratio is:

\frac{1 \text{ mol } CaCO_3}{2 \text{ mol } HCl}

2. Confusing Molarity with Moles

The Mistake:

Treating 0.75 M HCl as if it means 0.75 moles of HCl are present.

Why It Happens:

Molarity ( $M$ ) is moles per litre, not the total number of moles.

How to Avoid:

Always convert volume to litres first, then use the formula:

\text{Moles of HCl} = M \times V(\text{in L})

Here:

0.75 \text{ mol/L} \times 0.025 \text{ L} = 0.01875 \text{ mol HCl}

3. Unit Conversion Errors (mL → L)

The Mistake:

Using 25 mL directly as 25 L, or incorrectly converting (e.g., dividing by 100 instead of 1000).

How to Avoid:

Memorise: 1 L = 1000 mL. Always convert mL to L by dividing by 1000.

25 \text{ mL} = \frac{25}{1000} = 0.025 \text{ L}

4. Using Wrong Molar Mass Values

The Mistake:

Using atomic masses incorrectly — e.g., taking Ca as 40 g/mol but forgetting to add C and O₃.

How to Avoid:

Calculate molar mass step by step:

M_{CaCO_3} = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = \mathbf{100 \text{ g/mol}}

Write each element's contribution separately to avoid missing any.

--- …

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Q.Calcium carbonate reacts with aqueous HCl to give $CaCl_2$ and $CO_2$ according to the reaction, $CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2

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: Mole-to-Mass Stoichiometry via Molarity

Step-by-step solution

Common Mistakes in Molecular Mass Calculation Problems (HCl & CaCO₃ Reaction)

1. Forgetting the Mole Ratio from the Balanced Equation

2. Confusing Molarity with Moles

3. Unit Conversion Errors (mL → L)

4. Using Wrong Molar Mass Values

Q.Calcium carbonate reacts with aqueous HCl to give CaCl2CaCl_2CaCl2​ and CO2CO_2CO2​ according to the reaction, $CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2

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Q.Calcium carbonate reacts with aqueous HCl to give $CaCl_2$ and $CO_2$ according to the reaction, $CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2

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