Percent compositions

Weight/weight (w/w) percentage Weight/volume (w/v) percentage Volume/volume (v/v) percentage 

Unfortunately, although the percentage of solute is often listed, the method (w/w, w/v, v/v) is not. In this case, 1 normally assume that the method is weight/weight, but I m sure you imow about assumptions. 

Most of the solutions 1 talk about in the following examples of these percentages are aqueous solutions, solutions in which water is the solvent. 

for example, you dissolve 5.0 grams of sodium chloride in 45 grams of water, the weight percent is 

Suppose that you want to make 350.0 grams of a 5 percent (w/w) sucrose, or table sugar, solution. You know that 5 percent of the weight of the solution is sugar, so you can multiply the 350.0 grams by 0.05 to get the weight of the sugar  

The rest of the solution (350.0 grams - 17.5 grams = 332.5 grcims) is water. You can simply weigh out 17.5 grams of sugar and add it to 332.5 grams of water to get your 5 percent (w/w) solution. 

Chemists are often concerned with precisely what percentage of a compound s mass consists of one particular element. Lying awake at night, uttering prayers to Avogadro, they fret over this quantity, called percent composition. Calculating percent composition is trickier than you may think. Consider the following problem, for example. 

The human body is composed of 60 to 70 percent water, and water contains twice as many hydrogen atoms as oxygen atoms. If two-thirds of every water molecule is hydrogen and if water makes up 60 percent of the body, it seems logical to conclude that hydrogen makes up 40 percent of the body. Yet hydrogen is only the third most abundant element in the body by mass. What gives  

Oxygen is 16 times more massive than hydrogen, so equating atoms of hydrogen and atoms of oxygen is a bit like equating a toddler to a sumo wrestler. When the doors of the elevator won t close, the sumo wrestler is the first one you should kick out, weep though he may. 

Within a compound, it s important to sort out the atomic toddlers from the atomic sumo wrestlers. To do so, follow three simple steps  

Calculate the molar mass of the compound, as we explain in the preceding section. 

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Percent of theoretical O2 absorbed Weight percent composition in reaction mixture ... 

This corresponds to a distribution of 66% anti and 34% gauche. Table 3.2 gives the relationship between free-energy difference, equilibrium constant, and percent composition of a two-component mixture. 

Substrate Base, solvent Percent composition of alkene ... 

Start with a basis of 1 lb-mole of the natural gas at T = 80°F = 540°R and P = 40 psig = 54.7 psia. The volume percent and mole percent compositions are identical for a perfect-gas mixture. 

Click Coached Problems for a self-study module on percent composition. 

The percent composition of a compound is specified by citing the mass percents of the elements present. For example, in a 100-g sample of water there are 11.19 g of hydrogen and 88.81 g of oxygen. Hence the percentages of the two elements are... 

Use the formula of a compound to find percent composition or its equivalent (Example 3.4 Problems 33-42) 34,36... 

An alloy made up of tin and copper is prepared by simultaneously electroplating the two metals from a solution containing Sn(N03)2 and Cu(N03)2. If 20.0% of the total current is used to plate tin, while 80.0% is used to plate copper, what is the percent composition of tire alloy ... 

A 0.500-g sample of zinc-copper alloy was treated with dilute hydrochloric acid. The hydrogen gas evolved was collected by water displacement at 27°C and a total pressure of 755 mm Hg. The volume of the water displaced by the gas is 105.7 mL. What is the percent composition, by mass, erf the alloy (Vapor pressure of H20 at 27°C is 26.74 mm Hg.) Assume only the zinc reacts. 

Percent composition Mass percents of the elements in a compound, 56-57... 

The mass percent composition is a listing of the mass of each element present in 100 g of a compound. This percentage by mass listing is also called the compound s elemental analysis. 

We illustrate how the mass percent composition of a compound is related to its chemical formula using ammonium nitrate (NH4 NO3). The molar masses of NH4 NO3 and its constituent elements can be used to convert the chemical formula into mass percentages. 

The elemental analysis of a compound is usually determined by a laboratory that specializes in this technique. A chemist who has prepared a new compound sends a sample to the laboratory for analysis. The laboratory charges a fee that depends on the type and number of elements analyzed. The results are returned to the chemist as a listing of mass percent composition. The chemist must then figure out which chemical formula matches this composition. If a chemist has reason to expect a particular chemical formula, the observed percentages can be matched against the calculated percentages for the expected formula. This process is illustrated in Example 3-13. 

Because each chemical element is conserved, the masses of the products are equal to the masses of the elements contained in the original compound. This lets us complete the elemental analysis of the compound. The 5.00-g sample contained 4.63 g mercury and 0.37 g oxygen. The percent composition is found by dividing each elemental mass by the total mass and multiplying by 100 ... 

C03-0081. Portland cement contains CaO, Si02, AI2 O3, and FC2 O3. Calculate the mass percent composition of each compound. 

C03-0112. Nickel sulfate hexahydrate is NiSOq 6 H2 O. (a) Compute its molar mass, (b) Compute the number of moles contained in 25.0 g of this compound, (c) Determine its percent composition, (d) Determine the mass of this compound that contains 1.00 mol of O. 

C03-0125. A 3.75-g sample of compound that contains sulfur and fluorine contains 2.93 g of fluorine. The molar mass Is less than 200 g/mol. Calculate the percent composition of the compound and determine its molecular formula. 

Atoms and their symbols were introduced in Chap. 3 and 1. In this chapter, the representation of compounds by their formulas will be developed. The formula for a compound (Sec. 4.3) contains much information of use to the chemist. We will learn how to calculate the number of atoms of each element in a formula unit of a compound. Since atoms are so tiny, we will learn to use large groups of atoms—moles of atoms—to ease our calculations. We will learn to calculate the percent by mass of each element in the compound. We will learn how to calculate the simplest formula from percent composition data, and to calculate molecular formulas from simplest formulas and molecular weights. The procedure for writing formulas from names or from knowledge of the elements involved will be presented in Chaps. 5. ft. and 13. 

EXAMPLE 4.10. Calculate the percent composition of MgS04 that is, the percent by mass of each element in the compound. 

In laboratory work, the identity of a compound may be established by determining its percent composition experimentally and then comparing the results with the percent composition calculated from its formula. 

The first step is to determine the empirical formula from the percent composition data. 

Calculate the percent composition of each of the following (a) C4HN and (b) ChHl2. 

The percentages are the same as those in part (a). That result might have been expected. Since the ratio of atoms of carbon to atoms of hydrogen is the same (1 2) in both compounds, the ratio of masses also ought to be the same, and their percent by mass ought to be the same. From another viewpoint, this result means that the two compounds cannot be distinguished from each other by their percent compositions alone. 

Since the percentages are the same, the drug could be tetrahydrocannabinol. (It is not proven to be, however. If the percent composition were different, it would be proven not to be pure tetrahydrocannabinol.)... 

Which one of the following could possibly be defined as the ratio of moles of each of the given elements to moles of each of the others (a) Empirical formula, (b) molecular formula, or (c) percent composition by mass. 

Ans. Choice (a). This is a useful definition of empirical formula. The molecular formula gives the ratio of moles of each element to moles of the compound, plus the information given by the empirical formula. The percent composition does not deal with moles, but is a ratio of masses. 

Calculate the percent composition of C2H4. (h) Calculate the percent composition of C4HK. (c) Compare the results and explain the reason for these results. 

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