Mole Calculations: Mastering Body Parts Ratios for Efficient Problem Solving

Homework 6 posted and due Monday 3/5 Sigh, another Monday…

Moles Level 2: `Body Parts’ Calculations- the Process 1) Convert any mass or count to moles (divide up) 2) Set up appropriate ratio with unknown mol quantity (m) and solve for m. 3) If necessary re-convert m to mass or count (multiply down)

Mole Calculations: part 2 (cont.) 6) moles to atom count how many atoms of O are present in 0.8333 mol of C6 H12O6? =30*1023

7) Moles to atom count: how many atoms of C are in 0.02777 mol of C6H12O6? (1 mol =6*1023) • 1.662*1022 • 1.3*1024 • 1.0*1023 • 1.0*1024

Mole Calculations: part 2 (cont.) 8) mass to molecules: how many atoms of H in C6 H12O6are combined with 84 g of C ? 84 *1023 atoms ofH 8) atoms to mass: how many grams of H are combined with 2.4*1024 atoms of O in C6H12O 6 ? 8 grams H

9) mass molecules: How many molecules of C6H12O6 are present in a sample containing 32 g of O (at. wt=16 g/mol)? 1 1 mol count=6*1023 • 12*1023 • 5*1024 • 3.3*1023 • 2.0*1023

Mole Calculations: part 2 (cont.) 10) atoms to mass: how many grams of H are combined with 2.4*1024 atoms of O in C6H12O 6 ? 8 grams H

How many grams of C are in a sample containing 4.1666*1022 molecules of C6H12O6 . (1 mol count=6*1023) • 0.833 g • 5.000 g • 10.37 g • 17.28 g

The road trip through mole land so far… • Basic mole-mass-count conversions ( divide up, multiply down): =Moles level 1 • Mole ratio conversion within a compound (`body parts’ relationships) = Moles level 2

The mole road ahead... • % composition problems and combustion analysis (pp. 94-103) : moles level 2b • Reaction balancing (pp. 105-108) • Reaction stoichiometry predictions, limiting yields and % yields ( pp. 108-123): moles level 3

Mole Calculations: Mastering Body Parts Ratios for Efficient Problem Solving