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# Stoichiometry Formulas

The quantitative study of the relative amounts of reactants and products in chemical reactions is referred to as stoichiometry. Stoichiometric problems are solved in just four simple steps which are as following:

• First step is the balance the equation
• Convert units of a given reactants and products into moles.
• This is followed by calculating the moles of substance yielded by the reaction using the mole ratio,
• The final step is to convert moles of desired substance to desired units.
Let's suppose the following reaction of iron and oxygen that result in rust formation for stoichiometry calculation:

$Fe + O_2 \rightarrow Fe_{2}O_{3}$

As mentioned, the first step in stoichiometric calculation is the balancing the equation. As we know that reactants of a chemical equation are never destroyed or lost and therefore, the products of a reaction exactly correspond to the original reactants. Now, let’s balance the given equation:

$Fe + O_2 \rightarrow Fe_{2}O_{3}$

According to the equation, 1 iron (Fe) atom react with two oxygen (O) atoms to obtain 2 iron atoms and 3 oxygen atoms; the reaction is not balanced.

$4Fe + 3O_{2} \rightarrow 2Fe_{2}O_{3}$

According to the above mentioned balanced equation, 4 atoms of iron and 6 atoms of oxygen (since 3 $\times$ 2 = 6 ) react to form 4 iron (since 2 $\times$ 2 = 4 ) and 6 oxygen (2 $\times$ 3 = 6). The atoms on both sides of the equation match. Now, convert the given units to moles using conversion factors and then calculate the moles of substance.

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## Stoichiometry Formulas List

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Stoichiometry formulas used for stoichiometry calculations are as follows:

• One mole (1 mol) of an element = 6.022  $\times$ 10$^{23}$ entities (the Avogadro Number).
Therefore, 1 mol of carbon-12 means there are 6.022 $\times$ 10$^{23}$ carbon-12 atoms; 1 mol of H$_{2}$O contains 6.022  $\times$ 10$^{23}$ H$_{2}$O molecules.
• Molar mass of an element = mass(g)/mole.
• Molar mass of an compound = Sum total of molar masses of atoms of elements in the formula
• Mole to mass stoichiometry formula:  Mass (g) = $\frac{no. \ of \ moles \times no. \ of \ grams}{1 \ mol}$
• Mass to mole stoichiometry formula: No. of moles = $\frac{Mass (g) \times \ 1 \ mol}{no \ of \ grams}$
• Moles to substance entity (atoms, molecules etc) formula: No. of entities = $\frac{no. \ of \ moles \times \ 6.022 \times 10^{23} \ entities}{1 \ mol}$
• Substance entity to mole stoichiometry formula: No. of moles = $\frac{no. \ of \ entities \times 1 \ mol}{6.022 \times 10^{23} \ entities}$
• Mass percent of an element in a compound is calculated using molecular mass and chemical formula.
The formula is as follows:

Mass % of element X = $\frac{atoms \ of \ X \ in \ formula \times atomic \ mass \ of \ X \ (amu)}{molecular (or formula) \ mass \ of \ compound \ (amu)}$ $\times$ 100

Since the chemical formula provides information about moles of each element in a compound, the mass percent of element X can be determined using the molar mass. The stoichiometry formula is as follows

Mass % of element X = $\frac{moles \ of \ X \ in \ formula \times molar \ mass \ of \ X \ (amu)}{mass \ of \ one \ mole \ of \ compound (amu)}$ $\times$ 100.

## Stoichiometry Formulas Mole to Mole

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The mole to mole stoichiometry formulas determine the molar relationships of reactants and products in a chemical reaction using the balanced chemical equation which in turn is used to calculate the moles of a product produced from the given quantities of the reactants. It follows the law of conservation of mass which states that reactants of a chemical equation are never destroyed or lost and therefore, the products of a reaction exactly correspond to the original reactants. Using the law, the given equation is balanced and the stoichiometric ratios of the reactants and products; these ratios can be used as conversion factors for mole-to-mole conversions.

Now, let’s suppose, we want to calculate the number of mole produced by 2 moles of oxygen. For the purpose, the following equation will be used
$H_{2} + O_{2} \rightarrow H_{2} O$.

Now, let’s first balance the equation. By balancing the equation, we get following one.

$2H_{2} + O_{2} \rightarrow H_{2} O$.

Here, we can see one mole of oxygen produces two moles of water. The stoichiometric ratio is 1 mole of oxygen: 2 mole of water. Let’s assume that there is abundant hydrogen gas available for reaction. Under this condition,

$\frac{2 \ mole \ of \ oxygen \times 2 \ mole \ of \ water}{1 \ mole \ of \ oxygen}$ = 4 moles of water

## Mass to Mole Stoichiometry Formula

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It uses molar mass as a conversion ration. Molar mass refers to the mass of a given substance divided by its amount (mol). As we know that mass of a substance is calculated by measuring the amount of the substance and then converting it into the moles. This is followed by calculating the substance's molar mass by multiplying its relative atomic mass to the molar mass constant (1 g/mol).The molar mass constant is then multiplied by the given mass to convert mass to moles.

Example: For example, convert 20 grams of water to moles of water. The molar mass of water is 20 g/mol. Therefore the molecular weight of water in g/mol can be used to convert gram to mole. Thus,

$\frac{20 \ g \ water \times 1 \ mol}{20 g \ water}$ = 1 mol water.

## Grams to Grams Stoichiometry Formula

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The grams to grams stoichiometry formula converts mass of one substance into that of other which cannot be done directly. Instead, it is done by converting the mass of first substance into mole using mass to mole stoichiometry formula followed by mole to mole conversion factors and then convert the mole of second substance into mass. The first step is to use molar mass as conversion factor to convert the mass of substance A into mole. The second step is to use mole ratio as conversion factor to convert mole of substance A into that of substance B. Third step is to use molar ratio to convert the mole of substance B into mass of substance B.

## Gas Stoichiometry Formula

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According to the “ideal gas law”, at temperature (T) in Kelvin, the volume (V) occupied by n moles of any gas has a pressure (P).

The equation for gas law is given as following:
P V = n R T.

Here R is known as the gas constant and the equation is called the ideal gas law or equation of state. Under these conditions, 1 mole of gas occupies 22.4 litres of volume. The ideal gas law and balanced gas equations are used for stoichiometry calculations of gaseous reactions using gas stoichiometry formula.
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