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Stoichiometry

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For any chemical change there’s both quantitative and qualitative aspect for both reactants and products. The stoichiometry looks into the quantitative aspect of chemical change for both reactants and products. 

The stoichiometry actually looks into lots of things before the calculations are carried out. The molar ratio, molar masses and inter conversion of these into specific mole numbers are part of stoichiometry. We can also say that stoichiometry looks into the broader aspect of any chemical reactions.

Stoichiometry Definition

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The quantitative study of any chemical change is considered under stoichiometry. This also takes into account the molar ratio of both reactants and products along with their molar mass. The inter conversion of mass of substance present and changing those to their individual moles is the basic strategy.

Apart from these the stoichiometry of any chemical change begins with balancing of chemical reaction. Once the balancing is over, the required number of moles are calculated from the ratio and mass of the substances present in initial stage.

Conversion or molar proportion is very much essential for understanding the stoichiometric aspect of chemical reaction. In some cases the molar number is calculated back to their individual mass. 

Percent Yield of a Reaction

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Percent yield of a chemical change is the yield that is calculated from the actual mass produced and compared to theoretical yield during the chemical change. For every chemical reaction there’s an ideal mass expected where the balanced reaction is taken into consideration. 

In some specific chemical changes the conditions and actual mass produced is calculated from the balanced reaction. The percent yield is then calculated from actual mass yield and theoretical yield.

Percent yield = [(actual yield / theoretical yield)] * 100

The values are never found to be over 100 percent as under theoretical conditions that possibility is rare.

Molecular Velocity of Gas

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The molecular velocity of gas taken into consideration by calculating two specific aspect of the gaseous molecule. The temperature of the system, as it increases the randomness of gaseous molecule. Next, the molecular mass of the gaseous molecule as it becomes a decisive factor in the final velocity of the gaseous molecule.

The root mean square velocity = √ (3 RT / M)

Here, R stands for universal gas constant, T is temperature in Kelvin scale and M is molecular mass of the gaseous molecule.

Percent Composition by Mass

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The percent composition takes into account the molar mass of the compound and then followed by the individual atomic mass of the participants. In order to get the individual atom’s percent, the molar mass of the compound is taken into account. This is followed by specific atomic mass calculations (atomicity is taken into account wherever necessary). 

Percent composition = [(atomic mass / molar mass)] * 100

For example, the molar mass of calcium carbonate (CaCO3) = 100 g
Ca = (40 / 100) * 100 = 40%
O = (48/100) * 100 = 48%
C = (12/100) * 100 = 12%

Ideal Gas Law Stoichiometry

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The gas law which is applied for all ideal gases is termed as ideal gas law and the universal gas constant is taken into consideration based on what units are used for calculations.

P V = n R T 

Here, P is pressure taken into account from either atmosphere point of view or Torricelli, Pascal etc.
The ‘n’ is the number of moles which goes into the reaction.
R is the universal gas constant.
T is temperature given out either in Kelvin or Centigrade.

Stoichiometry Steps

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For a general stoichiometry problem, the first step is to balance the chemical reaction. Once the balancing is carried out, the individual moles are taken into account. Depending upon the requirement, either we have to convert mass given into mole numbers or moles into mass of substance. 

The solid materials we have to go and find the molality in case that is essential or else for liquid substances we need to use the molarity to figure out the concentration. The volume of gaseous materials are also calculated based on the number of moles that we get from balancing of chemical reaction. 

Stoichiometry Problems

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Find out the mass number of water that we get by reacting 80 grams of Oxygen with corresponding hydrogen molecule.

Solution: 
Step I
Write the balanced reaction of water formation.
2 H2 + O2  2 H2O 
From the balanced reaction, we can see 2 moles of hydrogen gas reacting with 1 mole of oxygen gas. The product that comes out is 2 moles of water.
Step II
Comparative mole for hydrogen is 10 as the original ratio between H2 and O2 was 2: 1 
Step III
The 10 moles of hydrogen corresponds with 5 moles of oxygen which is again equivalent to 10 moles of water.
Step IV
5 moles of Oxygen turns into 5 * 32 = 160 g 
10 moles of H2O turns into 10 * 18 = 180 g