Stoichiometry

Stoichiometry is a branch of chemistry which uses relationships between what reacts and what is formed in a chemical reaction to obtain required data quantitatively.

Before studying about how to perform calculations in stoichiometry, it is important to know some terms so that the calculation can be easy.

The percentage weight of an element present in the formula weight of a chemical compound is called percentage composition. In a way, percentage composition is a part of each constituent element present in \(100\) parts of a chemical compound:

\[\begin{align} \text {(Percentage composition)} &= \dfrac {(\text {Weight of an element in 1-gram molecule of the compound})}{(\text {Gram molecular weight of the compound})} × 100\\ \\\\ &=\dfrac {(\text {Weight of an element in 1-gram molecule of the compound})}{(\text {Mole weight of the compound})} × 100. \end{align}\]

Now, let us understand the steps to calculate the percent composition of an element in a compound.

• Step 1: Calculate gram molecular weight of the compound.
• Step 2: Calculate the weight of the given element in the formula of a compound.
• Step 3: Calculate the percentage of the weight of the given element in the weight of the compound.

An example may make the steps of calculation more clear.

Calculate the percentage of water of crystallisation in hydrated copper(II) sulphate: \(\big[\ce{CuSO_4.5H_2O}\big].\) \(\ \ (\ce{Cu}=64; \ce{S}=32; \ce{O}=16; \ce{H}=1)\)

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We have that

• \(g\)-molecular weight of \(\ce{CuSO_4.5H_2O}\) is equal to \(1(\ce{Cu}) + 1(\ce{S}) + 4(\ce{O}) + 5(\ce{H_2O}) = 1(64) + 1(32) + 4(16) + 5(18) = 64+32+64+90 = 250\text{ (g)};\)
• \(g\)-molecular weight of water of crystallisation is equal to \(5(18) = 90\text{ (g)}.\)

So, the percentage of water of crystallisation in \(\ce{CuSO_4.5H_2O}\) is \(\frac {90}{250}×100=36\%\). \(_\square\)

Let us try another example, but in this case, we are taking it to the next level.

Calculate the percentage of pure iron in \(10\text{ kg}\) of Iron(II) oxide \(\big[\ce{Fe_2O_3}\big]\) of \(80%\) purity. \(\ (\ce{Fe}=56; \ce{O}=16)\)

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The weight of pure \(\ce{Fe_2O_3}\) in impure \( \ce{Fe_2O_3}\) is \(10 × \dfrac {80}{100} = 8\text{ kg}=8000\text{ g}.\)

\(g\)-molecular weight of \( \ce{Fe_2O_3}\) is \(2(56)+3(16) = 160\text{ g}.\)

The weight of iron in 1-gram molecule of \( \ce{Fe_2O_3}\) is \(2(56) = 112\text{ g}.\)

Now, when pure \( \ce{Fe_2O_3}\) ore is \(160\text{ g}\), the weight of pure iron is \(112\text{ g}.\)

So, when pure \( \ce{Fe_2O_3}\) ore is \(8000\text{ g}\), the weight of pure iron is \(\dfrac{112×8}{160} = 5600\text{ g} = 5.6\text{ kg}.\)

So, the percentage of pure iron in \(10\text{ kg}\) of ore is \(\dfrac {5.6}{10} × 100 = 56%\). \(_\square\)

Now that the basics of stoichiometry have been cleared, based on the above example, let us try another example.

Potassium nitrate \(\big[\ce{KNO_3}\big]\) and \(\big[\ce{K_3PO_4}\big]\) are two potash fertilizers. If both fertilizers cost the same, which of the two is better? \(\ (\ce{K}=39; \ce{N}=14; \ce{O}=16; \ce{P}=31)\)

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1. \(g\)-molecular weight of \(\ce{KNO_3}\) is \(1(39) + 1(14) + 3(16) = 101\text{ g}.\)
   The weight of ptassium in \(1\text{ g}\)-molecule of \(\ce{KNO_3}\) is \(39\text{ g}.\)
   Therefore, the percenatge composition of potassium in \(\ce{KNO_3}\) is \(\frac{39}{101} \times 100 = 38.61\).

2. \(g\)-molecular weight of \(\ce{K_3PO_4}\) is \(3(39) + 1(31) + 4(16) = 212\text{ g}.\)
   The weight of potassium in \(1\text{ g}\)-molecule of \(\ce{K_3PO_4}\) is \(3(39) = 117\text{ g}.\)
   Therefore, the percenatge composition of potassium in \(\ce{K_3PO_4}\) is \(\frac{117}{212} \times 100 = 55.18\).

So, potassium phosphate is a better fertilizer as it contains more percentage of potassium as compared to potassium nitrate. \(_\square\)

Let us now revise what molecular and empirical formulas are.

Molecular formula: The chemical formula of a substance which represents the actual number of atoms of each element present in one molecule of a compound is called the molecular compound.

Example:

Empirical formula: It is the formula of a compound which shows the simplest whole number ratio between the atoms of various elements in the compound.

Steps for determination of empirical formula of a compound from its percentage composition:

Now, we move on to the second part of stoichiometry. Let us first see the steps of calculation.

• Step 1. Write the percentage weight (or weight) and atomic weight of each element present in a given compound in a table.
• Step 2. Divide the percentage weight (or weight) of each element with atomic weight and find the correct answer to two decimal places. The ratio gives the number of moles (g-atoms) of each element in the compound.
• Step 3. Divide each ratio of the relative number of moles obtained in step 2 by the smallest whole number ratio. This gives the simplest ratio of atoms present in a compound. If the ratios obtained are not whole numbers, then multiply them by the smallest suitable integer so that all the ratios are whole numbers.
• Step 4. Write down the empirical formula showing atoms of various elements in a simple ratio of whole numbers.

A diagrammatic summary of calculation of empirical formula

Now, let us try an example to make the steps clear.

A colored hydrated salt has the following percentage composition:

\[\ce{Cu}=25.60\%,\quad \ce{S}=12.8\%,\quad \ce{O}= 25.60\%,\quad \text{Water of crystallization } (\ce{H_2O}) = 36\%.\]

Calculate the empirical formula of the hydrated salt. \(\ (\ce{Cu}=64 ; \ce{S}= 32 ; \ce{O}=16 ; \ce{H}=1)\)

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We have the following table:

Element	Percentage weight	Atomic weight	Relative number of moles	Simple ratio of the atoms
\(\ce{Cu}\)	\(25.60\)	\(64\)	\(25.60 ÷ 64 = 0.40\)	\(0.40 ÷ 0.40 = 1\)
\(\ce{S}\)	\(12.80\)	\(32\)	\(12.80 ÷ 32 = 0.40\)	\(0.40 ÷ 0.40 = 1\)
\(\ce{Cu}\)	\(25.60\)	\(16\)	\(25.60 ÷ 16 = 1.60\)	\(1.60 ÷ 0.40 = 4\)
\(\ce{Cu}\)	\(36.00\)	\(18\)	\(36.00 ÷ 18= 2.00\)	\(2.00 ÷ 0.40 = 5\)

So, the empirical formula of the compound is \(\ce{CuSO_4.5H_2O}\). \(_\square\)

Determination of molecular formula of a compound from its percentage composition:

Now that we have understood how to calculate the empirical formula, we are going to learn how to calculate the molecular formula from the empirical formula and vapor density. Let us first see the steps of calculation:

• Step 1. Calculate the empirical formula as explained above.
• Step 2. Calculate the empirical weight from the empirical formula.
• Step 3. Calculate molecular weight from the vapor density.
• Step 4. Calculate the value of \(n\), the ratio of the molecular weight to the empirical formula weight.
• Step 5. Calculate the molecular formula by multiplying \(n\) with empirical formula weight.

Let us see an example to make the steps clear.

A gaseous hydrocarbon of vapor density 29 contains 82.76% of carbon. Calculate its molecular formula. \([\ce{C}=12 ; \ce{H}=1]\)

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We have the following table:

Element	Percentage weight	Atomic weight	Relative number of moles	Simple ratio of the atoms
\(\ce{C}\)	\(82.76\)	\(12\)	\(82.76 ÷ 12 = 6.89\)	\(6.89 ÷ 6.89= 1\) (or) \(1 × 2 = 2\)
\(\ce{H}\)	\(100-82.76=17.24\)	\(1\)	\(17.24 ÷ 1 = 17.24\)	\(17.24÷6.89= 2.5\) (or) \(2.5 × 2 =5\)

So, the empirical formula of the hydrocarbon is \(\ce{C_2H_5}.\)
The empirical formula weight is \((2×12)+(1×5)= 24+5=29.\)
Also, the molecular weight of hydrocarbon is \(2×\text{V.D.} = 2×29 = 58.\)
Thus, \(n\) (whole number positive integer) =\(\dfrac{\text{Molecular weight}}{\text{Empirical formula weight}}\) =\(\dfrac {58}{29} = 2.\)

So, the molecular formula is \(n × (\text{Empirical formula})\) = \(2 × \ce{C_2H_5} = \ce{C_4H_{10}}\). \(_\square\)

Great! Let's now finish off the explanation with an excellent example:

The empirical formula of a compound is \(\ce{PQ}\). If its empirical formula weight is \(\frac {1}{3}\) of its vapor density, calculate the molecular formula of the compound.

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The molecular formula is \(n×\dfrac {\text {Molecular formula weight}}{\text {Empirical formula weight}}
= \dfrac{2×\text{V.D.}}{\text {Empirical formula weight}}.\)

Now, the empirical formula weight is \(\frac {1}{3} ×\text{V.D.}\)

Substituting the value in the main equation, we get \(n = \frac {2×\text{V.D.}}{\frac {1}{3} × \text{V.D.}} = 6.\)

So, the molecular formula is \(n(\ce{PQ})= 6(\ce{PQ}) = \ce{P_6Q_6}\). \(_\square\)

Since you have now mastered this topic, try to solve these problems.