Question

CHEM 1405 - Experiment 7: Reaction Stoichiometry 15 Name: Section: Date: Report Sheet: Pre-lab Questions (pay attention to significant figures) Q1. Complete the following table. Name Formula No. of atoms of each element represented in the formula Mass of one mole of the substance (6x12.01 + 12 x1.008 + 6 *16.00) g = 180.16 g Glucose CH1206 6 atoms of C, 12 atoms of H, and 6 atoms of O Potassium sulfate Calcium phosphate Ethyl alcohol CH3OH Q2. Given: The molar mass of glucose is 180.16 g/mole. Avogadro's number is 6.02 x 10"! (a) What is the mass of 2.50 moles of glucose? (b) How many moles of glucose are in 6.48 g of glucose? (C) How many glucose molecules are in 2.50 moles of glucose?

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Answer 1

The solution of question 2 is as follows

2)the molar mass of glucose = 180-16 g/mole which means 1 mole of glucose weighs = 180 16 G :2.50 moles of glucose weighs - 180.16 x2.50 = 450.4g (B) No. of moles of given mass of glucose Molar mass of glucose glucose given mass of glucose= 6.48 g Molar mass of glucose = 180.16 g/more No. of moles of glucose in 6.489 180.16 g/more - 6.48 g sample = 0.035 moie. 0.035 moles 04 glucose is present in. 6.48 g of glucose.

(() 1 more of glucose contains = 6:0 23X1023 no oy molecules 2.50 moles of glucose contains 2.50 X 6.023x1023 no. oy glucose molecules - 15.058 1023 No. of glucose molecules

The answer of question 1 is as follows

1) Name Formula 1No. of atoms mass of of each element one more represented in loy substance the formula 6 atoms oy C, 12 Otom (6x12.01 +12 3008 Glucose C6H. Of of H and 6 atoms +6x1600) g. of o potasium 12 atoms of K. one (2x39 + 1x32 K₂ Soy Tatom of s u atom - 1749 Sulfate oyo (3x40+2x31 - Calcium 13 atoms of ca, a 1 Phosphate caz (Poy atoms of p, 8 atong +8x169 of o = 310g 1 = 180.169 +4X164 Ethyl alcohol 2 atoms of my 12x12 + 6x1 CH₂OHC, 6 atoms of tх16) g. H one atom = 4.69 of o 2H5OH :6