University:
Miami University
Course:
CHM 141 | College Chemistry
Academic year:

2020
Views:

193

Pages:

2
Author:

Alma Stephenson

Name: __________________________________ Period: ______________________ Date: ________ Chemistry Formula Triangles Student Lab Answers How to use a formula triangle: 1. Write down the term you wish to calculate (e.g. n =). 2. Cover the term you wish to calculate with your forefinger. 3. The remain two terms are those you must use in your equation on the other side of the = sign, working from the top down and/or left to right, and include all multiplication/division signs linking the two remaining terms. Practice Problems: 1. Calculate the following: a. The molarity of a solution where 1.9 moles of iron (II) chloride are dissolved to make 1750 mL of solution. n = 1.9 V = 1750ml (1.75L) C=n/V C = 1.9/1.75 C = 1.09 mol/L b. The molarity of a solution where 1.1 moles of silver nitrate are dissolved to make 0.250L of solution. n = 1.1 V = 0.250L C=n/V C = 1.1/0.250 C = 4.4 mol/L c. The number of moles of Na2CO3 in 0.750L of solution if the concentration is 0.640mol/L? V = 0.750l C= 0.640 mol/L n=CxV n = 0.640 x 0.750 n = 0.48 mol d. The molarity if 3.2 moles of CaCl2 are dissolved to make 1000 ml of solution? n = 3.2 V=1 C=n/V C = 3.2/1 C = 3.2 mol/L e. The volume of 1.7 moles of sucrose made into a 3.4mol/l solution. n = 1.7 C = 3.4 mol/L V = n/C V = 1.7/3.4 Name: __________________________________ Period: ______________________ Date: ________ Chemistry Formula Triangles Student Lab Answers V = 0.5L f. The number of moles of 54.9 grams in K2SO4. K2SO4 Mr = (39.098 x 2) + 32.06 + (15.999 x 4) =174.252 m = 54.9 n = m/Mr n = 54.9/174.252 n = 0.32 mol g. The mass of 11 mol of hydrogen chloride. HCl Mr = 1.008 + 35.45 = 36.458 n = 11 mol m = n x Mr m = 36.458 x 11 m = 401.04g h. The number of moles of 99.4 grams of NaCl. NaCl Mr = 22.90 + 35.45 = 58.35 m = 99.4 n = m/Mr n = 99.4/58.35 n = 1.7 mol 2. Expert Level (hint you must use both equations) a. The molarity when 54.8 grams of lithium sulfate are dissolved to make 250 mL of solution. Li2SO4 Mr = (6.94 x 2) + 32.06 + (15.999 x 4) =109.936 m = 54.8g n = m/Mr n = 54.8 / 109.936 n = 0.5 mol C = n/V C = 0.5 / 0.250L C = 2 mol/L b. The molarity when 99.1 grams of (NH4)2SO4 are dissolved to make 0.5 L of solution. (NH4)2SO4 Mr = (14.007 + (1.008 x4) x2) + 32.06 + (15.999 x 4) =132.134 m = 99.1g n = m/Mr n = 99.1 / 132.134 n = 0.75 mol C = n/V C = 0.75 / 0.50L C = 1.5 mol/L