The principles of stoichiometry give a fair idea about the various methods applied for mathematical steps involving chemical reactions where the masses of reactants and products and/or their respective volumes are calculated.

The laws of conservation of matter, conservation of energy and combining weights in chemical reactions constitute the basis of stoichiometric considerations. The applications of these laws to industrial processes is not obvious to the technical worker.

In a chemical change, matter is neither created nor destroyed, but only changed from one form to another. This law requires that

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Anyone, working with chemicals, should know the behavior of substances entering into or resulting from a chemical change along with the relative amounts of theses chemicals which react with one another and the amount of products formed as a result of chemical change.

The calculation of relative amounts of substances in chemical reactions is called Stoichiometry.

The calculation of relative amounts of substances in chemical reactions is called Stoichiometry.

The percentage measurements in chemistry deal with the same underlying concept. The percentage is about checking how many would we have out of hundred. In most of the calculations we make use of the quantity of the provided mass or volume of the molecule or compound and then divide it by the accepted mass or volume of one mole of the molecule or compound.

Percent yield is calculated in an equivalent way by subtracting the actual yield from the theoretical yield and then dividing the theoretical yield. Percent composition is also quite similar, though much simpler. Divide the number of grams of each particular atom in one mole of a substance by its molar mass and we have its percent composition. For all these quantities remember that the answer will be in decimal form so we will need to multiply it by 100 to change it into percentage form.

**Example Problem**

If 11.2g of Na reacts with 9.00g of H_{2}O and 0.400g of H_{2} is formed, what is the percent yield of the reaction?

**Solution:**

The chemical equation for this problem can be written as

**2Na + 2H**_{2}O →** 2NaOH + H**_{2}

__Tips__

Key clues that indicate percent yield problems are

**a)** Amount of both reactants given

**b)** An amount for the product

**c)** The words percent yield

In percent yield problems, limiting reactant calculations are frequently performed first to determine the theoretical yields. This is the amount of product that is formed if the reaction goes 100% to completion (this rarely happens in the lab) and what was calculated in exercises 4 and 5.

**% of yield =** $\frac{Actual\ yield}{Theoritical\ yield}$** $\times 100$**

= $\frac{0.400}{0.494}$**$\times 100$ = 81.0 %**

Percent yield is calculated in an equivalent way by subtracting the actual yield from the theoretical yield and then dividing the theoretical yield. Percent composition is also quite similar, though much simpler. Divide the number of grams of each particular atom in one mole of a substance by its molar mass and we have its percent composition. For all these quantities remember that the answer will be in decimal form so we will need to multiply it by 100 to change it into percentage form.

If 11.2g of Na reacts with 9.00g of H

Key clues that indicate percent yield problems are

In percent yield problems, limiting reactant calculations are frequently performed first to determine the theoretical yields. This is the amount of product that is formed if the reaction goes 100% to completion (this rarely happens in the lab) and what was calculated in exercises 4 and 5.

= $\frac{0.400}{0.494}$

According to Maxwell, the distribution of molecular speeds at a particular temperature is given by the following curve.

- Fraction of molecules with too high and too low speeds are very small.
- No molecules have zero velocity.
- Initially the fractions of molecules increase with increase in velocity, reaches the peak of the curve (corresponding to most probable velocity) and thereafter decreases with increase in velocity.

The percent by mass of each element present in a compound is called its elemental analysis. Elemental analysis, also called the mass percent composition, is a catalog of the mass of each element present in 100g of a compound.

To show how the mass percent composition of a compound is related to its chemical formula we begin with a compound whose formula is already known. As mentioned in the previous section, ammonium nitrate (NH_{4}NO_{3}) is an important fertilizer. The molar masses of NH_{4}NO_{3 }and its constituent elements can be used to convert the chemical formula into mass percentages.

The formula tells how many moles of each element are present in 1 mol of the substance. One mole of NH_{4}NO_{3} contains 2 mol of nitrogen 4 mol of hydrogen and 3 mol of oxygen. The masses of the element contained in 1 mol follows.

**2 mol N x 14.007g / mol = 28.014g N**

4 mol H x 1.008g / mol = 4.032g H

3 mol O x 15.999g / mol = 47.997g O

Mass of 1 mol NH_{4}NO_{3} = 80.043g

Each mole of NH_{4}NO_{3 }has a total mass of 80.043g. Of this total, 28.014g is nitrogen. The ratio of these is the mass fraction of nitrogen and the percent nitrogen by mass is the mass fraction multiplied by 100.

(28.014g / 80.043g) = 0.35000 (mass fraction of nitrogen in NH_{4}NO_{3})

0.35000 x 100 = 35.000% (mass percent of nitrogen in NH_{4}NO_{3})

Many chemical reactions either consume or give off gases. The ideal gas law can be used to relate the volumes of such gases to the amounts of other substances involved in the reaction. The ideal gas equation allows chemists to work stoichiometry problems involving gases. For example, the decomposition of potassium chlorate to potassium chloride and oxygen by heating.To show how the mass percent composition of a compound is related to its chemical formula we begin with a compound whose formula is already known. As mentioned in the previous section, ammonium nitrate (NH

The formula tells how many moles of each element are present in 1 mol of the substance. One mole of NH

4 mol H x 1.008g / mol = 4.032g H

3 mol O x 15.999g / mol = 47.997g O

Mass of 1 mol NH

Each mole of NH

(28.014g / 80.043g) = 0.35000 (mass fraction of nitrogen in NH

0.35000 x 100 = 35.000% (mass percent of nitrogen in NH

This problem can be solved in three steps

Calculate the number of moles of oxygen gas produced.

$\frac{25.0_{g} KClO_{3}}{1}$ * $\frac{1\ mol\ O_{2}}{122.55_{g} KClO_{3}}$ * $\frac{3\ mol\ O_{2}}{2\ mol\ KClO_{2}}$ = 0.3060 mol O

Convert the temperature to kelvin and pressure to atmosphere

700 torr / 760 torr/atm = 0.9211 atm

Put everything in ideal gas equation

PV = nRT

v = $\frac{nRT}{p}$

= $\frac{(0.3060\ mol)(0.0821 \frac{L\ atm}{K\ atm})(300\ K)}{0.9211\ atm}$

= 8.18 L

Where, $V_{m}$ = Molar volume at 0

The mean molar mass of an ideal gas mixture is equal to the sum of the products of volume percentage and molar mass of each individual gas.

With standard temperature and pressure of a gas mixture,

Where $\rho_{mn}$

To raise a given natural number to an indicated power, the logarithm of the number is multiplied by the exponent indicating the power; the antilogarithm of the result is the required power.

To extract a required root of a given number divide the logarithm of the number by the index to its root; the antilogarithm of the quotient is the required root.

To multiply natural numbers, add their logarithms and the antilogarithm of their sum is the required product.

To divide natural numbers subtract the logarithm of the divisor from the logarithm of the dividend, the antilogarithm of the difference is the required quotient.

Since this is not a tutorial on logarithms, our discussion has been brief and elementary, being restricted to the need of the present occasion. The interested reader may find in almost any intermediate algebra, trigonometry or business mathematics book, a clear explanation of common logarithm, with adequate tables.

It is obvious that tables of logarithms may be calculated to various degrees of approximation, they may be calculated to 5, 6, 7 or a higher number of decimal places.

Mass fraction for a given element = $\frac{Mass\ of\ the\ element\ present\ in\ 1\ mole\ of\ compound}{Mass\ of\ 1\ mole\ of\ compound}$

Steps for calculation of percentage composition of a compound.

**Step 1-**Calculate the weight of the element in one molecule of the compound.**Step 2 -**Calculate the molecular mass of the compound by adding up the atomic weight of each element.**Step 3 -**The percentage of an element is calculated by the relation

Answer- Mass of Boron in a mole of borax = 4 x 11g = 44g

Molecular mass of borax = 2 x 23 + 4 x 11 + 7 x 16 + 10 x 18 = 382

Percentage of boron = (44g/382g) x 100 = 11.5%

Answer - Molar mass of Na

Mass of water = 10 x 18 = 180g

Percentage of water of crystallization = (180g/286g) x100

= 62.93%

To solve the problems based on chemical equations we need to keep the following points in mind.

- Write down balanced equation in molecular form.
- Underline those substances whose weights/volume have been mentioned in the numerical or whose weights/volumes have to be determined.
- Write the atomic weights/moles/molar volume of the underlined substances depending upon the nature of the problem.
- Write down the actual quantities of the substances which had been underlined. For those substances whose weights/volumes have been calculated just write the sign of interrogation (?).
- Calculate the result by unitary method.

In this type of problem, mass of one of the reactants/products is to be calculated if that of the other reactants/products is given.

In this type of problem, mass or volume of one of the reactants or products is calculated from the mass or volume of the other substances. While solving such problems, we should make use of the fundamentals that one mole of each gas occupies a volume of 22.4L at STP.

In this type of problem, the volume of one of the reactants or products is to be calculated from the volume of some other reactants or products.

How many grams and how many moles of SiF

The balanced chemical equation for the reaction is

- Step 1 SiO
_{2}+ 4HF → SiF_{4}+ 2H_{2}O - Step 2 4 moles 1 mole
- Step 3 4(1+19) 28.1+ 4x19

- Step 4 63.4g ?

Molecular mass of HF= 20; 20g HF = 1 mole ; 63.4g HF = 3.17 mole

According to the balanced chemical equation,

4moles of HF produce SiF4 = 1mole

3.17 moles HF produce SiF4 = 3.17/4 =

Molecular mass of SIF4 = 104.1; 1mole of SiF4= 104.1g

0.793 mole of SiF4 = 104.1x0.793 g =

The balanced chemical equation involved is

NaCl + AgNO_{3} → AgCl + NaNO_{3}

1mole 1 mole 23+35.5 108+35.5

= 58.5g = 143.5g

? 14.53g

It is evident from the above equation:

143.5g of AgCl is obtained from 58.5g NaCl

14.35g of AgCl is obtained from 58.5 x 14.35/143.5 = 5.85g NaCl

Percentage of purity (5.85/6.5) x 100% = 90%

The balanced chemical equation involved in the burning of hydrogen gas is

2H_{2} + O_{2} → 2H_{2}O

2moles 1 mole2x2 =4g 32g

2x22.4 22.4L

= 44.8L

56 L ?

From the above data, it is evident that:

44.8 L of H

56 L of H

The chemical equation involved in the production of O

Now 22.4L of O

28L of O

The balanced chemical equation involved is:

1mole 1 mole

100g 22.4L at STP

From the above data, it is evident that

100g of marble gives 22.4L of CO

1g of marble gives 22.4/100 L = 0.224 L of CO

Let the formula of the hydrocarbon is C

The balanced chemical equation for the hydrocarbons combustion is as follows:

1 vol. (x+y/4) vol. x vol.

According to the data given, x+y/4 = 5; 4x + y = 20

Contraction in volume on shaking with caustic potash solution (KOH) = volume of evolved CO

Therefore, x=3.

Substituting the value of x in equation 4x + y = 20 we will get y=8.

The formula of hydrocarbon is C

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