Formula mole ratios

The calculations in Example 3.4 illustrate an important characteristic of formulas. In one mole of Fe203 there are two moles of Fe (111.7 g) and three moles of O (48.00 g). This is the same as the atom ratio in Fe203,2 atoms Fe 3 atoms O. In general, the subscripts in a formula represent not only the atom ratio in which the different dements are combined but also the mole ratio. 

The strategy used to calculate the simplest formula depends to some extent on which of these types of information is given. The basic objective in each case is to find the number of moles of each element, then the simplest mole ratio, and finally the simplest formula. 

Strategy First (1), convert the masses of the three elements to moles. Knowing the number of moles (n) of K, Cr, and O, you can then (2) calculate the mole ratios. Finally (3), equate the mole ratio to the atom ratio, which gives you the simplest formula. 

Reality Check A mole ratio of 1.00 A 1.00 B 333 C would imply a formula A3B3C10 if the mole ratio were 1.00 A 2.50 B 5.50 C, the formula would be A2B5Cn. In general, multiply through by the smallest whole number that will give integers for all the subscripts. 

Find the mole ratios and then the simplest formula. 

STRATEGY We convert from the given volume of gas into moles of molecules (by using the molar volume), then into moles of reactant molecules or formula units (by using a mole ratio), and then into the mass of reactant (by using its molar mass). If the molar volume at the stated conditions is not available, then use the ideal gas law to calculate the amount of gas molecules. 

The mole ratio is 1.50mol Sn/3.()0mol O = lmol Sn/2mol O. The formula is Sn02. 

If each of the following mole ratios is obtained in an empirical formula problem, what should it be multiplied by to get an integer ratio ... 

A chemical equation describes a chemical reaction in many ways as an empirical formula describes a chemical compound. The equation describes not only which substances react, but the relative number of moles of each undergoing reaction and the relative number of moles of each product formed. Note especially that it is the mole ratios in which the substances react, not how much is present, that the equation describes. In order to show the quantitative relationships, the equation must be balanced. That is, it must have the same number of atoms of each element used up and produced (except for special equations that describe nuclear reactions). The law of conservation of mass is thus obeyed, and also the "law of conservation of atoms. Coefficients are used before the formulas for elements and compounds to tell how many formula units of that substance are involved in the reaction. A coefficient does not imply any chemical bonding between units of the substance it is placed before. The number of atoms involved in each formula unit is multiplied by the coefficient to get the total number of atoms of each element involved. Later, when equations with individual ions are written (Chap. 9), the net charge on each side of the equation, as well as the numbers of atoms of each element, must be the same to have a balanced equation. The absence of a coefficient in a balanced equation implies a coefficient of 1. 

Ans. The empirical formula can be determined from the ratio of moles of carbon atoms to moles of hydrogen atoms. From the masses of products, we can get the numbers of moles of products, from which we can get the numbers of moles of C and H. The mole ratio of these two elements is the same in the products as it is in the original compound. 

Zinc borates with different mole ratios of Zn0 B203 H20 can be readily prepared by reacting zinc oxide with boric acid (Fig. 1). Among all these zinc borates, the compound with a molecular formula of 2Zn0-3B203-3.5H20, is the most commonly used fire retardant. This article will review recent research results on the use of this particular form of zinc borate as a multifunctional fire retardant. 

The rules of stoichiometry also apply in this case. In electrochemical cells, we must consider not only the stoichiometry related to chemical formulas, but also the stoichiometry related to electric currents. The half-reaction under consideration not only involves 1 mol of each of the copper species, but also 2 mol of electrons. We can construct a mole ratio that includes moles of electrons or we could construct a mole ratio using faradays. A faradav (F) is a mole of electrons. Thus, we could use either of the following ratios for the copper half-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 same mole ratio, a/d, can also be found by simply balancing the common element in the formulas of A and D. Thus, the ratio, a/d, is the same as the ratio QS/QK seen in the equation for a gravimetric factor derived in Section 3.6.3 (Equation (3.12)) and is used to convert the weight of one substance to the weight of another, just as we described in Chapter 3 was the purpose of the gravimetric factor. Thus the concept of a gravimetric factor is based on stoichiometry. 

Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Order the elements according to the general rules for naming ionic and molecular compounds. 

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. 

The C H mole ratio of 1 2.26 means that we can write CiH2 2g as a temporary formula for the liquid. Multiplying the subscripts by small integers in a trial-and-error procedure until whole numbers are found then gives the final formula. In the present instance, multiplication of the subscripts by 4 is needed ... 

Just as we can derive the empirical formula of a substance from its percent composition, we can also calculate the percent composition of a substance from its empirical (or molecular) formula. The strategies for the two kinds of calculations are exactly opposite. Aspirin, for example, has the molecular formula C9H8O4 and thus has a CH 0 mole ratio of 9 8 4. We can convert this mole ratio into a mass ratio, and thus into percent composition, by carrying out mole-to-gram conversions. 

The percent composition of glucose can be calculated either from the molecular formula (CgH Og) or from the empirical formula (CH20). Using the molecular formula, for instance, the C H 0 mole ratio of 6 12 6 can be converted into a mass ratio by assuming that we have 1 mol of compound and carrying out mole-to-gram conversions ... 

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 law of multiple proportions, in its historical form, is somewhat contusing, but we can understand it better by determining the empirical formulas of the nitrogen oxides. Following the procedure of Table 2-2, we convert the masses of N and O contained in 100 g of each compound to moles as shown in Table 2-5. The mole ratios then indicate the relative numbers of N and O atoms in one molecule ... 

Column (5) is determined in this problem, as in the previous problem, by dividing both numbers by the smallest (0.0559) which preserves the mole ratio of 0.0559 0.391 and leads to the whole-number ratio required to write a chemical formula. Column (5) contains 6.99, a number that is so close to a whole number that the difference can be taken for experimental error. The mole ratio of Na2SC>4 to H2O is 1 to 7, providing us with an empirical formula of Na2S04- 7H20. 

Section 3.1) (Bush and Trager, 1981). This was achieved by working with the complete ion cluster of the molecule and inverting the mole ratio. Starting from the formula... 

Also, since the ratios are the same in a mole as in a molecule, the molecular formula can be used to obtain mole ratios of elements in a compound. 

From the weight of CuCl2, and the weight of Cu, subtraction will give the weight of Cl. From these weights, the mole ratio of copper to chlorine, the empirical formula, and the percent composition of CuCl2 can then be calculated. 

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