How to calculate the mole ratio of a compound

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How to calculate the mole ratio of a compound

Calculate the molar ratio between two substances given their balanced reaction Molar ratios state the proportions of reactants and products that are used and formed in a chemical reaction. Molar ratios can be derived from the coefficients of a balanced chemical equation. Stoichiometric

coefficients of a balanced equation and molar ratios do not tell the actual amounts of reactants consumed and products formed. Chemical equations are symbolic representations of chemical reactions. In a chemical equation, the reacting materials are written on the left, and the products are

written on the right; the two sides are usually separated by an arrow showing the direction of the reaction. The numerical coefficient next to each entity denotes the absolute stoichiometric amount used in the reaction. Because the law of conservation of mass dictates that the quantity of each

element must remain unchanged over the course of a chemical reaction, each side of a balanced chemical equation must have the same quantity of each particular element. In a balanced chemical equation, the coefficients can be used to determine the relative amount of molecules, formula

units, or moles of compounds that participate in the reaction. The coefficients in a balanced equation can be used as molar ratios, which can act as conversion factors to relate the reactants to the products. These conversion factors state the ratio of reactants that react but do not tell exactly

how much of each substance is actually involved in the reaction. Determining Molar Ratios The molar ratios identify how many moles of product are formed from a certain amount of reactant, as well as the number of moles of a reactant needed to completely react with a certain amount of

another reactant. For example, look at this equation: [latex]CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O[/latex] From this reaction equation, it is possible to deduce the following molar ratios: 1 mol CH4: 1 mol CO2 1 mol CH4: 2 mol H2O 1 mol CH4: 2 mol O2 2 mol O2: 1 mol CO2 2 mol

O2: 2 mol H2O In other words, 1 mol of methane will produced 1 mole of carbon dioxide (as long as the reaction goes to completion and there is plenty of oxygen present). These molar ratios can also be expressed as fractions. For example, 1 mol CH4: 1 mol CO2 can be expressed as

[latex]\frac{1 \ mol \ CH_4}{1 \ mol \ CO_2}[/latex]. These molar ratios will be very important for quantitative chemistry calculations that will be discussed in later concepts. In a chemical reaction, compounds react in a set ratio. If the ratio is unbalanced, there will be leftover reactant. To

understand this, you need to be familiar with the molar ratio or mole ratio. A mole ratio is ?the ratio between the amounts in moles of any two compounds involved in a chemical reaction. Mole ratios are used as conversion factors between products and reactants in many chemistry problems.

The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation. Also known as: The mole ratio is also called the mole-to-mole ratio. For the reaction:2 H2(g) + O2(g) ¡ú 2 H2O(g) The mole ratio between O2 and H2O is 1:2. For every 1

mole of O2 used, 2 moles of H2O are formed. The mole ratio between H2 and H2O is 1:1. For every 2 moles of H2 used, 2 moles of H2O are formed. If 4 moles of hydrogen were used, then 4 moles of water would be produced. For another example, let's start with an unbalanced equation:

O3 ¡ú O2 By inspection, you can see this equation is not balanced because mass is not conserved. There are more oxygen atoms in ozone (O3) than there are in oxygen gas (O2). You cannot calculate mole ratio for an unbalanced equation. Balancing this equation yields: 2O3 ¡ú 3O2 Now

you can use the coefficients in front of ozone and oxygen to find the mole ratio. The ratio is 2 ozone to 3 oxygen, or 2:3. How do you use this? Let's say you are asked to find how many grams of oxygen are produced when you react 0.2 grams of ozone. The first step is to find how many

moles of ozone are in 0.2 grams. (Remember, it's a molar ratio, so in most equations, the ratio is not the same for grams.)To convert grams to moles, look up the atomic weight of oxygen on the periodic table. There are 16.00 grams of oxygen per mole.To find how many moles there are in

0.2 grams, solve for:x moles = 0.2 grams * (1 mole/16.00 grams).You get 0.0125 moles.Use the mole ratio to find how many moles of oxygen are produced by 0.0125 moles of ozone:moles of oxygen = 0.0125 moles ozone * (3 moles oxygen/2 moles ozone).Solving for this, you get 0.01875

moles of oxygen gas.Finally, convert? the number of moles of oxygen gas into grams for the answer:grams of oxygen gas = 0.01875 moles * (16.00 grams/mole)grams of oxygen gas = 0.3 grams It should be fairly obvious that you could have plugged in the mole fraction right away in this

particular example because only one type of atom was present on both sides of the equation. However, it's good to know the procedure for when you come across more complicated problems to solve. Updated November 05, 2018 By Chris Deziel In stoichiometry, or the study of relative

amounts of substances in reactions, you'll come across two situations that call for the calculation of mole ratio. In one, you're analyzing a mystery substance to determine its empirical formula, and in the other, you're calculating relative amounts of reactants and products in a reaction. In the

first case, you usually have to weigh the individual components of a compound and calculate the number of moles of each. In the second case, you can usually find the mole ratio by balancing the equation for the reaction. The typical procedure to determine the empirical formula of a

mystery compound is to analyze it for its component elements. If you obtain the weight of each element in the compound, you can determine the number of moles of each compound by dividing the actual weight in grams by the atomic weight of that element. To do this, you have to look up

the atomic weights in the periodic table or, to make things easier on yourself, you can use an online mole calculator that automatically converts between weight in grams and number of moles. Once you know the number of moles of each component of the compound, you divide each by the

one with the lowest number and round to the nearest integer. The numbers are the mole ratios, and they appear as subscripts in the empirical formula. Example: You analyze a compound and find that it contains 0.675 g of hydrogen (H), 10.8 g of oxygen (O) and 13.5 g of calcium (Ca).

What is the empirical formula? The molar mass of hydrogen is 1 g (rounding to one decimal place), so the number of moles present in the compound is 0.675/1 = 0.675. The molar mass of oxygen is 16 g, and the molar mass of calcium is 40.1 g. Performing the same operation for these

elements, you find that the number of moles of each element are: H ¨C 0.675 O ¨C 0.675 Ca ¨C 0.337 Calcium is the element with the lowest number of moles, which is 0.337. Divide this number into the others to obtain the mole ratio. In this case, it's H ¨C 2, O ¨C 2 and Ca ¨C 1. In other words, for

every calcium atom in the compound, there are two hydrogens and two oxygens. The numbers derived as the mole ratio of the elements appear in the empirical formula as subscripts. The empirical formula for the compound is CaO2H2, which is usually written Ca (OH)2. If you know the

reactants and products of a reaction, you can write an unbalanced equation for the reaction by putting the reactants on one side and the products on the other. The law of conservation of mass requires that both sides of the equation must have the same number of atoms of each element,

and this provides the clue on how to find the mole ratio. Multiply each side of the equation by a factor that balances the equation. The multiplication factors appear as coefficients, and these coefficients tell you the mole ratios of each of the compounds in the reaction. For example, hydrogen

and oxygen combine to form water. The unbalanced equation is H2 + O2 ¨C> H2O. However, this equation isn't balanced because there are more oxygen atoms on one side than the other. The balanced equation is 2H2 + O2 ¨C> 2 H2O. It takes two hydrogen atoms for every oxygen atom to

produce this reaction, so the mole ratio between hydrogen and oxygen is 2:1. The reaction produces two water molecules, so the mole ratio between oxygen and water is 1:2, but the mole ratio between water and hydrogen is 2:2. ?You want to add some sections to the porch. Before you go

to the hardware store to buy lumber, you need to determine the unit composition (the material between two large uprights). You count how many posts, how many boards, how many rails - then you decide how many sections you want to add before you calculate the amount of building

material needed for your porch expansion. Stoichiometry problems can be characterized by two things: (1) the information given in the problem, and (2) the information that is to be solved for, referred to as the unknown. The given and the unknown may both be reactants, both be products,

or one may be a reactant while the other is a product. The amounts of the substances can be expressed in moles. However, in a laboratory situation, it is common to determine the amount of a substance by finding its mass in grams. The amount of a gaseous substance may be expressed

by its volume. In this concept, we will focus on the type of problem where both the given and the unknown quantities are expressed in moles. Figure \(\PageIndex{1}\): Mole ratio relationship. Chemical equations express the amounts of reactants and products in a reactants and products in a

reaction. The coefficients of a balanced equation can represent either the number of molecules or the number of moles of each substance. The production of ammonia \(\left( \ce{NH_3} \right)\) from nitrogen and hydrogen gases is an important industrial reaction called the Haber process,

after German chemist Fritz Haber. \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)\] The balanced equation can be analyzed in several ways, as shown in the figure below. Figure \(\PageIndex{2}\): This representation of the production of ammonia

from nitrogen and hydrogen show several ways to interpret the quantitative information of a chemical reaction. We see that 1 molecule of nitrogen reacts with 3 molecules of hydrogen to form 2 molecules of ammonia. This is the smallest possible relative amounts of the reactants and

products. To consider larger relative amounts, each coefficient can be multiplied by the same number. For example, 10 molecules of nitrogen would react with 30 molecules of hydrogen to produce 20 molecules of ammonia. The most useful quantity for counting particles is the mole. So if

each coefficient is multiplied by a mole, the balanced chemical equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any balanced chemical equation. Finally, if each mole quantity is converted to

grams by using the molar mass, we can see that the law of conservation of mass is followed. \(1 \: \ce{mol}\) of nitrogen has a mass of \(28.02 \: \text{g}\), while \(3 \: \text{mol}\) of hydrogen has a mass of \(6.06 \: \text{g}\), and \(2 \: \text{mol}\) of ammonia has a mass of \(34.08 \: \text{g}\). \

[28.02 \: \text{g} \: \ce{N_2} + 6.06 \: \text{g} \: \ce{H_2} \rightarrow 34.08 \: \text{g} \: \ce{NH_3}\] Mass and the number of atoms must be conserved in any chemical reaction. The number of molecules is not necessarily conserved. Figure \(\PageIndex{3}\): Apparatus for running Haber

process. A mole ratio is a conversion factor that relates the amounts in moles of any two substances in a chemical reaction. The numbers in a conversion factor come from the coefficients of the balanced chemical equation. The following six mole ratios can be written for the ammonia

forming reaction above. \[\begin{array}{ccc} \dfrac{1 \: \text{mol} \: \ce{N_2}}{3 \: \text{mol} \: \ce{H_2}} & or & \dfrac{3 \: \text{mol} \: \ce{H_2}}{1 \: \text{mol} \: \ce{N_2}} \\ \dfrac{1 \: \text{mol} \: \ce{N_2}}{2 \: \text{mol} \: \ce{NH_3}} & or & \dfrac{2 \: \text{mol} \: \ce{NH_3}}{1 \: \text{mol} \:

\ce{N_2}} \\ \dfrac{3 \: \text{mol} \: \ce{H_2}}{2 \: \text{mol} \: \ce{NH_3}} & or & \dfrac{2 \: \text{mol} \: \ce{NH_3}}{3 \: \text{mol} \: \ce{H_2}} \end{array}\] In a mole ratio problem, the given substance, expressed in moles, is written first. The appropriate conversion factor is chosen in order to

convert from moles of the given substance to moles of the unknown. Example \(\PageIndex{1}\) How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen? Solution Step 1: List the known quantities and plan the problem. Known Given: \

(\ce{H_2} = 4.20 \: \text{mol}\) Unknown \(\text{mol}\) of \(\ce{NH_3}\) The conversion is from \(\text{mol} \: \ce{H_2}\) to \(\text{mol} \: \ce{NH_3}\). The problem states that there is an excess of nitrogen, so we do not need to be concerned with any mole ratio involving \(\ce{N_2}\). Choose the

conversion factor that has the \(\ce{NH_3}\) in the numerator and the \(\ce{H_2}\) in the denominator. Step 2: Solve. \[4.20 \: \text{mol} \: \ce{H_2} \times \dfrac{2 \: \text{mol} \: \ce{NH_3}}{3 \: \text{mol} \: \ce{H_2}} = 2.80 \: \text{mol} \: \ce{NH_3}\] The reaction of \(4.20 \: \text{mol}\) of

hydrogen with excess nitrogen produces \(2.80 \: \text{mol}\) of ammonia. Step 3: Think about your result. The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation.

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