WS3_key - Chem 142 Name Worksheet 3 Stoichiometry Summer 2011 Section 1 Calculate the average atomic mass of Zirconium(Zr 90Zr(51.45 91Zr(11.27

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chem 142 Name ________________________________________ Worksheet 3: Stoichiometry Summer 2011 Section ________________ 1. Calculate the average atomic mass of Zirconium (Zr). 90Zr (51.45%), 91Zr (11.27%), 92Zr (17.17%), 94Zr (17.33%),96Zr (2.78%). "( 90amu ! 0.5145 ) + ( 91amu ! 0.1127 ) % $ ' average atomic mass = $ + ( 92amu ! 0.1717 ) + ( 94amu ! 0.1733) ' = 91.32amu $ + ( 96amu ! 0.0278 ) ' # & 2. Sucrose is the common sugar used in all homes, and chemical analysis tells us that the chemical composition is 42.14% carbon, 6.48% hydrogen and 51.46% oxygen. What is the molecular formula of sucrose if its molar mass is approximately 340 g/mol? Assume 100 g of sucrose. 1mol C moles C = 42.14g C ! = 3.508mol C 12.011g C 1mol H moles H = 6.48g H ! = 6.429mol H 1.008g H 1mol O moles O = 51.46g O ! = 3.216mol O 16.00g O empirical formula CH2O empirical equivalents = 340 g mol sucrose = 11 30 g mol empirical molecular formula C11H22O11 3. Determine the empirical and molecular formulas for a deadly nerve gas that gives the following mass percent analysis. C = 39.10% H = 7.67% O = 26.11% P = 16.82% F = 10.30% The molar mass is known to be 184.1 g/mol. Assume 100 g of compound. 1mol C moles C = 39.10g C ! = 3.255mol C 12.011g C Page 1 of 3 Chem 142 Summer 2011 moles H = 7.76g H ! 1mol H = 7.61mol H 1.008g H 1mol O moles O = 26.11g O ! = 1.632mol O 16.00g O 1mol P moles P = 16.82g P ! = 0.5431mol P 30.97g P 1mol F moles F = 10.30g F ! = 0.54211mol F 19.00g F empirical formula C6H14O3PF empirical equivalents = 184.1g mol compound =1 184.1g mol empirical molecular formula C6H14O3PF 4. Erythrose (M = 120 g/mol) is an important chemical compound used often as a starting material in chemical synthesis and contains carbon, hydrogen, and oxygen. Combustion analysis of a 700.0 mg sample yielded: 1.027 g CO2 and 0.4194 g H2O. Determine the molecular formula. moles C = 1.027g CO2 ! 12.011g C 1mol C ! = 0.02334mol C 44.011g CO2 12.011g C 2.016g H 1mol H moles H = 6.48g H 2O ! ! = 0.04656mol H 18.016g H 2O 1.008g H 1mol O moles O = ( 0.700g erythrose ! 0.2803g C ! 0.04693g H ) " = 0.02330mol O 16.00g O empirical formula CH2O empirical equivalents = 120 g mol sucrose =4 30 g mol empirical molecular formula C4H8O4 5. A combustion device was used to determine the empirical formula of a compound containing ONLY carbon, hydrogen and oxygen. A 0.6349 g sample of the unknown produced 1.603 g of CO2 and 0.2810 g H2O. Page 2 of 3 Chem 142 Summer 2011 Determine the empirical formula of the compound. Assuming the empirical and molecular formulas are the same, what is the balanced chemical reaction for the combustion of this liquid? moles C = 1.603g CO2 ! 12.011g C 1mol C ! = 0.03642mol C 44.011g CO2 12.011g C 2.016g H 1mol H moles H = 0.2810g H 2O ! ! = 0.03119mol H 18.016g H 2O 1.008g H 1mol O moles O = ( 0.700g erythrose ! 0.4374g C ! 0.03144g H ) " = 0.01038mol O 16.00g O empirical formula C3.5H3O Need whole numbers so multiply everything by 2 to get: C7H6O2 2C7H6O2 (l)+ 15O2 (g) 6H2O (g) + 14CO2 (g) Page 3 of 3 ...
View Full Document

This note was uploaded on 01/19/2012 for the course CHEM 142A taught by Professor Campbell during the Summer '11 term at University of Washington.

Ask a homework question - tutors are online