Chemical calculations mole ratios

Why learn to write mole ratios They are the key to calculations that are based on chemical equations. Using a balanced chemical equation, mole ratios derived from the equation, and a given amount of one of the reactants or products, you can calculate the amount of any other participant in the reaction. 

A stoichiometry calculation is thus essentially a three-step procedure in which 1) the weight of D is divided by its formula weight to get moles of D, 2) the moles of D are converted to the moles of A by multiplying by the mole ratio a/d, as found in the chemical equation, and 3) the moles of A are converted to grams of A by multiplying by the formula weight of A. 

The pipeted volume is converted to moles by multiplying the liters of solution by its molarity. The moles of titrant are determined using the mole ratio in the balanced chemical equation for the acid—base reaction. The molarity of the solution is calculated by dividing the moles of titrant by the liters of titrant used. 

Convert the masses of the reactants and products to moles using their molar masses. Using the mole ratios from the balanced chemical equation, it is possible to determine how much material should react or be produced. These calculated values can be compared to the observed values. 

By measuring the mass of iron that reacts and the mass of copper metal produced, you can calculate the ratio of moles of reactant to moles of product. This mole ratio can be compared to the ratio found in the balanced chemical equation. 

Comparing and Contrasting Compare the ratio of moles of iron to moles of copper from the balanced chemical equation to the mole ratio calculated using your data. 

The balanced chemical equation for a reaction is used to set up the conversion factor from one substance to another and that conversion factor, the mole ratio for the reaction, is applied to the moles given to calculate the moles required. 

Knowing a compound s percent composition makes it possible to calculate the compound s chemical formula. As shown in Figure 3.8, the strategy is to find the relative number of moles of each element in the compound and then use the numbers to establish the mole ratios of the elements. The mole ratios, in turn, give the subscripts in the chemical formula. 

With the masses of the carbon-containing product (C02) and hydrogen-containing product (H20) known, the strategy is to then calculate the number of moles of carbon and hydrogen in the products, from which we can find the C H mole ratio. This information, in turn, provides the chemical formula, as outlined by the flow diagram in Figure 3.9. 

The most important step to all of these calculations is the use of a value known as the mole ratio. The mole ratio is the ratio of moles of one substance to moles of second substance. It is determined by the ratios of the coefficients from the balanced chemical equation. The mole ratio is used in all conversions since it allows you to switch from values that describe the given substances to values that describe the unknown substance. To facilitate this process, there is another chart, Figure 12.2, that provides guidelines for solving most problems. In this first type of calculation, we will use the mole ratio to convert from units of moles of the given substance to moles of the unknown substance. We re going to omit the states of the reactants and products so that you can focus your attention on the coefficients. 

The same basic formulas for mole conversion with one substance are used to compare two substances in a chemical reaction. The only difference is that one substance must be converted to another using the mole ratio before the calculation can be completed. 

Write a balanced chemical equation for the formation of vanadium(V) oxide. Use the known mole ratio of vanadium to oxygen to calculate the unknown amount of oxygen. 

According to the balanced chemical equation, the ratio of lithium nitride to water is 1/3. The ratio of lithium nitride to water, based on the mole amounts calculated, is 0.14 0.32. Divide this ratio by 0.14 to get 1.0 2.3. For each mole of lithium nitride, there are only 2.3 mol water. However, 3 mol are required by stoichiometry. Therefore, water is the limiting reactant. 

Suppose that a solution of an acid reacts with a solution of a base. You can determine the concentration of one solution if you know the concentration of the other. (This assumes that the volumes of both are accurately measured.) Use the concentration and volume of one solution to determine the amount (in moles) of reactant that it contains. The balanced chemical equation for the reaction describes the mole ratio in which the compounds combine. In the following Sample Problems and Practice Problems, you will see how to do these calculations. 

I A formula gives the mole ratio of one element to another and the mole ratio of each element to 1 mol of compound. With atomic masses and a chemical formula, we can calculate mass ratios—for example, percent by mass—rather easily. 

Inst as compounds have definite ratios of elements, chemical reactions have definite ratios of reactants and products. Those ratios are used in Section 10.1 to calcnlate the number of moles of other substances in a reaction from the nnm-ber of moles of any one of the snbstances. Section 10.2 combines information from Section 10.1, Chapter 7, and elsewhere to explain how to calcnlate the mass of any substance involved in a reaction from the mass of another. Section 10.3 demonstrates how to work with qnantities in nnits other than moles or masses when finding quantities of reactants or prodncts. Section 10.4 shows how to calcnlate the quantities of snbstances involved in a reaction even if the quantities of reactants present are not in the mole ratio of the balanced equation. Section 10.5 covers the calculation of the percentage yield of a product from the actual yield and the theoretical yield, based on the amonnt(s) of reactant(s). Section 10.6 explains which of these types of calcnlations can and cannot be done with net ionic equations. 

The balanced chemical equation gives the mole ratios of all the substances in the reaction, just as an empirical formula gives the ratios of atoms of the elements in a compound. As with chemical formulas, these ratios can be used as factors in calculations involving any two of the substances. 

Net ionic equations (Chapter 9), like all other balanced chemical equations, give the mole ratios of reactants and products. Therefore, any calculations that require mole ratios may be done with net ionic equations as well as with total equations. However, a net ionic equation does not yield mass data directly because part of each soluble ionic compound is not given. For example, we can tell how many moles of silver ion are required to produce a certain number of moles of a product. 

Step 2 Start the second row by entering the moles of limiting quantity—the same as the number of moles of that substance in the first row, since all of the limiting quantity is used up. The other entries in the second row are calculated using the technique of Sec. 10.1. The entries in row 2 will always be in the same mole ratio as the coefficients in the balanced chemical equation. 

How do we use the mole in chemical calculations Recall that Avogadro s number is defined as the number of atoms in exactly 12 grams of 12C. Thus 12 grams of 12C contains 6.022 X 1023 atoms. Also, a 12.01-gram sample of natural carbon contains 6.022 X 1023 atoms (a mixture of 12C, 13C, and 14C atoms, with an average mass of 12.01). Since the ratio of the masses of the samples (12 g/12.01 g) is the same as the ratio of the masses of the individual components (12 amu/12.01 amu), the two samples contain the same number of components. 

If you know the mole ratio of the two reacting chemicals in solution, you can calculate the amount (the number of moles and thus the number of grams) of the solute in the solution of unknown concentration. 

FIC. 6 Proton nmr spectrum of 2-methyl-3-pentanol containing Eu(dpm)3 (60 MHz). Mole ratio of Eu(dpm)3 to alcohol = 1.0. Compare this spectrum to those shown in Fig. 5. Protons nearest the hydroxyl group are shifted most. Methyl groups are recorded at reduced spectrum amplitude. Note the large chemical shift difference between the two protons on C-4. The average conformation of the molecule is the one shown and was calculated from the equation on p. 219. 

You may be wondering why you need to learn to write mole ratios. As you will see in the next section, mole ratios are the key to calculations based upon a chemical equation. Suppose you know the amount of one reactant you will use in a chemical reaction. With the chemical equation and the mole ratios, you can calculate the amount of any other reactant in the equation and the maximum amount of product you can obtain. 

Recall that stoichiometry is the study of quantitative relationships between the amounts of reactants used and the amounts of products formed by a chemical reaction. What are the tools needed for stoichiometric calculations All stoichiometric calculations begin with a balanced chemical equation, which indicates relative amounts of the substances that react and the products that form. Mole ratios based on the balanced chemical equation are also needed. You learned to write mole ratios in Section 12.1. Finally, mass-to-mole conversions similar to those you learned about in Chapter 11 are required. 

The next step involves determining whether the two reactants are in the correct mole ratio as given in the balanced chemical equation. The coefficients in the balanced chemical equation indicate that four moles of chlorine are needed to react with one mole of sulfur. This 4 1 ratio from the equation must be compared with the actual ratio of the moles of available reactants just calculated above. To determine the actual ratio of moles, divide the available moles of chlorine by the available moles of sulfur. 

What was the source of any deviation from the mole ratio calculated from the chemical equation How could you improve your results ... 

Mole ratios are central to stoichiometric calculations. They are derived from the coefficients in a balanced chemical equation. To write mole ratios, the number of moles of each reactant and product is placed, in turn, in the numerator of the ratio with the moles of each other reactant and product placed in the denominator. 

Mole ratios used in the calculations are determined from the balanced chemical equation. 

As a shortcut to determining the limiting reactant, all you have to do is to calculate the mole ratio(s) of the reactants and compare each ratio with the corresponding ratio of the coefficients of the reactants in the chemical equation thus ... 

If more than two reactants are present, you have to use one reactant as the reference substance, calculate the mole ratios of the other reactants in the feed relative to the reference, make pairwise comparisons versus the analogous ratios in the chemical... 

To perform this calculation the number of moles of phosphorus available has been multiplied by a stoichiometric factor, a mole ratio factor relating moles of the required reactant to moles of the other reactant. The stoichiometric factor comes directly from the coefficients in the balanced chemical equation. This is the reason you must balance chemical equations before proceeding with calculations. Here the calculation shows that 0.0702 mol of Gig is required to react with aU the available phosphorus. 

To detemoine the concentration of a solution, a solution of known concentration and volume can be treated with the unknown solution until the mole ratio is exactly what is required by the balanced chemical equation. Then from the known volumes of both reactants, the concentration of the unknown can be calculated, fhis procedure is called titration. An indicator is used to tell when to slop the titration. Typically an indicator is a compound that is one color (or colorless) in an acidic solution and a second color in a basic solution. 

From the viewpoint of chemical equilibrium, reactions (6) to (9) in the above system can be neglected because of the smaller Kp. Reactions (3) - (5) can also be neglected since they are unable to compete against reactions (1) and (2) when the mole ratio of H2 to CH4 is very large. Reactions (1) and (2) are treated by set theory according to the procedure proposed in a previous paper (2). The calculated results are listed in Table 1. 

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