Chemical formula mole ratios from

The chemical formula for a compound gives the ratio of atoms of each element in the compound to atoms of every other element in the compound. It also gives the ratio of dozens of atoms of each element in the compound to dozens of atoms of every other element in the compound. Moreover, it gives the ratio of moles of atoms of each element in the compound to moles of atoms of every other element in the compound. For example, a given quantity of H2O has 2 mol of H atoms for every mole of O atoms, and a given quantity of CH4 has 1 mol of C atoms for every 4 mol of H atoms. The mole ratio from the formula can be used as a factor to convert from moles of any element in the formula to moles of any other element or to moles of the formula unit as a whole. In Figure 7.2, these additional conversions have been added to those already presented in Figure 7.1. 

Then, using the mole ratio from the chemical formula (XCI3), we can calculate the moles of X contained in... 

This is the type of problem that smdents get wrong if they use the dilution formula (Eq. 3.3) by mistake. The dilution formula does not include the mole ratio from a balanced chemical equation, and in this case, the 2 1 ratio would result in an error of a factor of 2 if you used Eq. 3.3. 

In the problem above, we determined the percentage data from the chemical formula. We can determine the empirical formula if we know the percent compositions of the various elements. The empirical formula tells us what elements are present in the compound and the simplest whole-number ratio of elements. The data may be in terms of percentage, or mass or even moles. However, the procedure is still the same—convert each element to moles, divide each by the smallest, and then use an appropriate multiplier if necessary. We can then determine the empirical formula mass. If we know the actual molecular mass, dividing the molecular formula mass by the empirical formula mass, gives an integer (rounded if needed) that we can multiply each of the subscripts in the empirical formula. This gives the molecular (actual) formula, which tells what elements are in the compound and the actual number of each. 

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. 

It is remarkable that no empirical mixture parameters and no experimental data are required to use the equation. The only parameters in the Flory-Huggins equation are the hard core volumes V, which are a pme-component property, and the atomic or group contribution values are found in standard compilations. Since the v/s are significant in the FH equation only in terms of their ratios, pure-liquid molar volumes are often used for V in place of hard core volumes. For solutions of polymers of the same chemical formula, molecular masses are legitimate substitutes for V , for the same reason. Thus the volume fractions ( ) can be substituted by mass fractions W . Either volume fraction or mass fraction is directly related to laboratory data. To avoid mole fractions, the activity tti from Equations (4.368) and (4.369) can be used to calculate by / = aj. ... 

Although Chapter 9 was full of questions that began with, How much.. .T we are not done with such questions yet. In Chapter 9, our questions focused on chemical formulas. For example, we answered such questions as, How much of the element vanadium can be obtained from 2.3 metric tons of the compound V2O5 The chemical formula V2O5 told us that there are two moles of vanadium, V, in each mole of V2O5. We used this molar ratio to convert from moles of V2O5 to moles of vanadium. 

The ratio of moles of P4O10 to moles of P (which came from the subscripts in the chemical formula, P4O10) provided the key conversion factor that allowed us to convert from units of phosphorus to units of tetraphosphorus decoxide. 

Conversions between mass, moles, and the number of particles are summarized in Figure 10.11. Note that molar mass and the inverse of molar mass are conversion factors between mass and number of moles. Avogadros number and its inverse are the conversion factors between moles and the number of representative particles. To convert between moles and the number of moles of atoms or ions contained in the compound, use the ratio of moles of atoms or ions to 1 mole of compound or its inverse, which are shown on the upward and downward arrows in Figure 10.11. These ratios are derived from the subscripts in the chemical formula. 

Chemical stoichiometry is the area of study that considers the quantities of materials in chemical formulas and equations. Quite simply, it is chemical arithmetic. The word itself is derived from stoicheion, the Greek word for element and metron, the Greek word for measure. When based on chemical formulas, stoichiometry is used to convert between mass and moles, to calculate the number of atoms, to calculate percent composition, and to interpret the mole ratios expressed in a chemical formula. Most topics in chemical arithmetic depend on the interpretation of balanced chemical equations. Mass/mole conversions, calculation of limiting reagent and percent yield, and various relationships among reactants and products are commonly included in this topic area. 

Strategy In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form the compound. Therefore, we need to convert from mass percent to moles in order to determine the empirical formula. If we assume an exactly 100 g sample of the compound, do we know the mass of each element in the compound How do we then convert from grams to moles ... 

With ratios such as these—which come from the chemical formula—we can directly determine the amounts of the constituent elements present in a given amount of a compound without having to calculate mass percent composition. For example, we calculate the number of moles of Cl in 38.5 mol of CCI2F2 as follows ... 

To use the basic chemical quantity— the mole—to make calculations convenient To determine the empirical formula from percent composition or other mass-ratio data... 

The coefficients in a balanced equation give the ratio of moles of each substance in the reaction to moles of any other substance. They also give the ratio of formula units of each substance to formula units of any other substance. The balanced chemical equation is the cornerstone from which we can calculate how much of one substance reacts with or is produced by a certain quantity of another substance (Chapter 10). 

Stoichiometry involves the calculation of quantities of any substances involved in a chemical reaction from the quantities of the other substances. The balanced equation gives the ratios of formula units of all the substances in a chemical reaction. It also gives the corresponding ratios of moles of the substances. These relationships are shown in Figure 10.1. For example, one reaction of phosphorus with chlorine gas is governed by the equation... 

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