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More commonly, the empirical or simplest formula of a compound is determined by finding the masses of the elements in the compound, then calculating their mole ratio. The mole ratio is determined by dividing the smaller number of moles into the larger number of moles, in order to simplify the ratio to a whole number. Using the masses of copper and chlorine that you measured in the experiment (see previous page), calculate the moles of copper and the moles of chlorine below. Use the molar mass of each element to convert from grams to moles: 1. Moles 0.00917 g/mdes) = copper grams to Moles І ток ов си = 63.59 200917g/moles 2. Moles chlorine = 0·032299 g/mobe 2 1.145g 35.45 g/mole c 1 mole e al = 35.45/ 3. Write the number of moles of each element as a subscript: Cu 0.583 9 63.359/le 0.032299glades CI 4. Divide the above formula's subscripts by the smaller of the two subscripts to obtain the simplest (or empirical) formula of copper chloride: Cu Cl ng/a

More commonly, the empirical or simplest formula of a compound is determined by finding the masses of the elements in the compound, then calculating their mole ratio. The mole ratio is determined by dividing the smaller number of moles into the larger number of moles, in order to simplify the ratio to a whole number. Using the masses of copper and chlorine that you measured in the experiment (see previous page), calculate the moles of copper and the moles of chlorine below. Use the molar mass of each element to convert from grams to moles: 1. Moles 0.00917 g/mdes) = copper grams to Moles І ток ов си = 63.59 200917g/moles 2. Moles chlorine = 0·032299 g/mobe 2 1.145g 35.45 g/mole c 1 mole e al = 35.45/ 3. Write the number of moles of each element as a subscript: Cu 0.583 9 63.359/le 0.032299glades CI 4. Divide the above formula's subscripts by the smaller of the two subscripts to obtain the simplest (or empirical) formula of copper chloride: Cu Cl ng/a

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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and this table shows More commonly, the empirical or simplest formula of a compound is determined by finding the masses of the elements in the compound, then calculating their mole ratio. The mole ratio is determined by dividing the smaller number of moles into the larger number of moles, in order to simplify the ratio to a whole number. Using the masses of copper and chlorine that you measured in the experiment (see previous page), calculate the moles of copper and the moles of chlorine below. Use the molar mass of each element to convert from grams to moles: grams to Moles 1. Moles copper 10.00917 g/mdes I mole of C = 7 , to me it =0. صالح 63.59 =0009|7g|moles closer to our data then the rest of the compants сотрать 2. Moles chlorine = 0·032299 g/mok 2 1.145g 35.45 g/mdle I mole e al = 35.45 1 3. Write the number of moles of each element as a subscript: Cu 0.583 9 63.359/mole 0.032299 glades Cl 4. Divide the above formula's subscripts by the smaller of the two subscripts to obtain the simplest (or empirical) formula of copper chloride: Cu Cl 5. Does the simplest formula from your mole ratio calculation agree with the formula predict- might account for the Ì gloe
Transcribed Image Text:and this table shows More commonly, the empirical or simplest formula of a compound is determined by finding the masses of the elements in the compound, then calculating their mole ratio. The mole ratio is determined by dividing the smaller number of moles into the larger number of moles, in order to simplify the ratio to a whole number. Using the masses of copper and chlorine that you measured in the experiment (see previous page), calculate the moles of copper and the moles of chlorine below. Use the molar mass of each element to convert from grams to moles: grams to Moles 1. Moles copper 10.00917 g/mdes I mole of C = 7 , to me it =0. صالح 63.59 =0009|7g|moles closer to our data then the rest of the compants сотрать 2. Moles chlorine = 0·032299 g/mok 2 1.145g 35.45 g/mdle I mole e al = 35.45 1 3. Write the number of moles of each element as a subscript: Cu 0.583 9 63.359/mole 0.032299 glades Cl 4. Divide the above formula's subscripts by the smaller of the two subscripts to obtain the simplest (or empirical) formula of copper chloride: Cu Cl 5. Does the simplest formula from your mole ratio calculation agree with the formula predict- might account for the Ì gloe
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