EXPERIMENT 2 Determination of a Chemical Formula Additional Review Material Rele

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EXPERIMENT 2 Determination of a Chemical Formula Additional Review Material Relevant sections in the text (chemistry for Engineering students, Ed., Brown, Holme): 3.2 3.5 2" Techniques (see Blackboard/uaboratory/Techniques): vacuum Filtration (You Tube) Background The law of definite proportions states that a compound ahways contains two or more elements combined in the same definite proportion by mass. The chemical formula that represents this relative number of atoms in a compound (the simplest mole ratio) is called the empirical formula. The chemical formula that gives the actual number of atoms in a compound is called the molecular formula. Consider for example, a compound composed of two atoms, H and o. We know that water consists of two atoms of hydrogen and a single oxygen. Its molecular formula is Hzo. Another compound composed of these same elements, but in different proportions, is hydrogen peroxide. The molecular formula for hydrogen peroxide is Hao, and the empirical formula, specifying the smallest whole-number ratio of atoms in the compound, is Ho In the laboratory, the empirical formula of a compound can be determined by measuring the mass of each component in a compound. In this experiment, the empirical formula of a solid compound with the gener formula cucl zHzo will be determined by measuring mass of H20 the chlorine (cy in the compound. The values x, y, and z represent integers which will establish the proper chemical formula. This ionic compound is called a hydrate. Hydrates have a specific number of water molecules associated with each formula unit. When writing these chemical formulas, the formula of he is followed by dot which the "weak" bond between the anhydride and the number of water molecules (also called water of hydration Hydrates and the corresponding to gain or compounds often have distinctly different physical and chemical properties and their ability lose water makes them versatile and useful in a variety of practical applications. To determine the percent composition of water in the unknown Cuec, zH20 compound, water of hydration is first removed by heating the sample. mass of H20 %H20 in compoun (00) Remaining cu.cl, salts then readily dissolve to form aqueous solutions of the constituent ions Cly (s) We can now use the oxidation-reduction (redox) reaction between copper ions and solid magnesium (Me) to convert cu to cu(s) Cu (aq) Mg(s) cu(s) Mg2 (aq) (2.2)

Explanation / Answer

1. Mg required

(i) moles of Cu = 0.678 g/63.546 g/mol = 0.0107 mol

(ii) mass of Mg required = 0.0107 mol x 24.305 g/mol = 0.26 g

(iii) 1.5 times Mg used = 0.39 g

2. mass of Cl = 1.588 - 0.355 - 0678

                      = 0.555 g

3. mass%

Cu = 0.678 x 100/1.588 =42.7%

Cl = 0.555 x 100/1.588 = 34.95%

H2O = 0.355 x 100/1.588 = 22.35%

4. 100 g sample

Cu = 42.7/63.546 = 0.67 mol

Cl = 34.95/35.45 = 1.0 mol

H2O = 22.35/18 = 1.24 mol

divide by smallest factor

H2O = 2

Cl = 2

Cu = 1

Empirical formula = CuCl2.2H2O

5. mass%

Cu = 42.7%

Cl = 34.95%

H2O = 22.35%

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