Question

3. Consider 1.00 mole of hydrogen gas, H2. How many dihydrogen molecules are present? How many hydrogen atoms are present? What is the mass of this sample? 4. Ethanol has a molecular formula of CH3CH2OH. a. What is the average mass of one molecule of ethanol? b. What is the mass of 1.000 moles of ethanol? c. What is the mass of 0.5623 moles of ethanol, CH3CH2OH? d. How many moles of are present in a 100.0 g sample of ethanol? e. How many moles of each element (C, H, and O) are present in a 100.0 g sample of ethanol? f. How many grams of each element (C, H, and O) are present in a 100.0 g sample of ethanol?

Answer 1

I mdle = 6.022 x 1023 So, number of dihydrogen molecule present en 1.60 mde = 16.022 x 10'] I molecule of H₂ contains 2 H atoms. So, number of hydrogen atoms present = 2x 6.022 x 1023-17.204 x1024 24 redar mass of H₂ = 2:016 gmolt So, mass of this sample (1.00 mol H₂) = 2.016g | 4. a) Average mass of one molecule of CH₃ CH₂ OH = (12.011+ 3x1.008 + 12.011 + 2x1-008+160 +1.008) 46.07 u . Molecular mass express in grams equal to the molar mass. So, mass of of ethand = 46079 is 1.000 moles c) Mass of 0.5623 moles = 0.5623 * 46.079 | 25.909

d) Moles present in 100.0 g sample of ethand = 100.og 46.07 g mol- = [2.171 mol two one e) One molecule of CH₂CH₂OH contains B C elements, six H elements and o element. So moles oq c present in 10000g = 2.171 mo E 2x 2.171 mor 1. = 4.342 mor moles of H present = 6x21171 mol = (13.026 mol moles of o present = 1 x 2.171 md) = 2.171 med 52:15 $) Mars of C = 4-342 mo? x 12.011 g/mol = Mass of H = 13:026 mx 1008 god-'= (13.139 mass of O = 2.171 mol x 1600g mari" = 134.749)