Compounds percent composition

If the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition). For example, one... 

Empirical formulas can be calculated from the percent composition of a compound. You have already learned that the subscript numbers in formulas can be read in terms of the number of moles of each element. In 1 mole of N2Os, there are 2 moles of N and 5 moles of O. Read the formula as N2moies05 moies. The reason this is mentioned again is because any data that allows the number of moles of each element in a compound to be known allows the calculation of the empirical formula of that compound. Percent composition data does this. Here s the step-by-step way it is done ... 

The percent composition by mass is the percent by mass of each element in a compound. Percent composition is obtained by dividing the mass of each element in 1 mole of the compound by the molar mass of the compound and multiplying by 100 percent. Mathematically, the percent composition of an element in a compound is expressed as... 

A compound contains 6.0 g of carbon and 1.0 g of hydrogen, and has a molar mass of 42.0 g/mol. What are the compounds percent composition, empirical formula, and molecular formula ... 

If the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition). For example, one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is the part divided by the whole times 100% (or simply parts per hundred), so we can represent the percent composition of carbon dioxide as follows ... 

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... 

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 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. 

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. 

B We use the same technique as before determine the mass of each element in a mole of the compound. Their sum is the molar mass of the compound. The percent composition is determined by comparing the mass of each element with the molar mass of the compound. 

If we know the formula of a compound, it is a simple task to determine the percent composition of each element present. For example, suppose you wanted the percentage carbon and hydrogen in methane, CH4. First, calculate the molecular mass of methane ... 

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. 

Making particle numbers manageable with Avogadro s number Converting between masses, mole counts, and volumes Dissecting compounds with percent composition Moving from percent composition to empirical and molecular formulas... 

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. 

Use the following formula to find the percent composition for each element in the compound ... 

What if you don t know the formula of a compound Chemists sometimes find themselves in this disconcerting scencirio. Instead of cursing Avogadro (or perhaps after doing so), they analyze samples of the frustrating unknown to identify the percent composition. From there, they calculate the ratios of different types of atoms in the compound. They express these ratios as an empirical formula, the lowest whole-number ratio of elements in a compound. 

The beauty of this little trick is that you conveniently gift yourself with the same number of grams of each elemental component as its contribution to the percent composition. For example, if you assume that you have 100 g of a compound composed of 60.3% magnesium and 39.7% oxygen, you know that you have 60.3 g of magnesium and 39.7 g of oxygen. (The only time you don t do this is if the problem specifically gives you the masses of each element present in the unknown compound.)... 

Calculate the empirical formula of a compound with a percent composition of 88.9% oxygen and 11.1% hydrogen. 

To determine a moleculcir formula, you must know the gram formula mass of the compound as well as the empirical formula (or enough information to calculate it yourself from the percent composition see the preceding section for details). With these tools in hand, calculating the molecular formula involves three steps ... 

A compound has a percent composition of 49.5% carbon, 5.2% hydrogen, 16.5% oxygen, and 28.8% nitrogen. The compound s gram molecular mass is 194.2 g/mol. What are the empirical and molecular formulas ... 

Whenever a new compound is made in the laboratory or found in nature, it must be analyzed to find what elements it contains and how much of each element is present—that is, to find its composition. The percent composition of a compound is expressed by identifying the elements present and giving the mass percent of each. For example, we might express the percent composition of a certain colorless liquid found in gasoline by saying that it contains 84.1% carbon and 15.9% hydrogen by mass. In other words, a 100.0 g sample of the compound contains 84.1 g of carbon atoms and 15.9 g of hydrogen atoms. 

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