chapter-03 - Have to “do experiment” to get actual...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEM 1331 CHAPTER 3 Definition of Mole 6.02 x 10 23 Definition of Molar Mass : 1 mole of a substance weights the atomic/molecular mass in grams Important Equations moles = grams / molar mass Molecules = 6.02 x 10 23 x moles Molarity molarity = moles of solute / liters of solution Stoichiometry BALANCED EQUATIONS Relate relative numbers of moles of substances: If 2 substances A and B have coefficients a and b: Moles of B = (b / a) x moles of A ROADMAP STRESSES RELATIONSHIPS Moles of A n A Moles of B n B Particles N A Particles N B grams of A g Volume of Solution, L grams of B g Volume of Solution, L n B = (b/a)n A M = n / L n = g / MM N = 6.02 x 10 23 n N = 6.02 x 10 23 n Limiting Reagents If given information about more than one reactant: (a) calculate moles of “B” for each reactant individually (b) smaller number of moles of B comes from limiting reactant (c) use this number to finish problem
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Percent Yield %yield = {[actual yield] / [theoretical yield]} x 100 Calculate theoretical yield as above
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Have to “do experiment” to get actual yield Empirical Formula Ratio of number of each type of atom Related to MOLECULAR FORMULA by whole number multiple Calculate from relative masses of elements For each element in substance: (a) convert to moles (b) convert moles to “nice” numbers by dividing by smallest number (c) if necessary, multiply ALL by 2, 3 etc to clear fractions If given PERCENT of each element, use percent as grams If not given one element, subtract other masses from total sum For combustion of C, H compounds or C, H, O compounds Sometimes given g of CO 2 and H 2 O, Convert to: g of C by multiplying CO 2 mass by (12 / 44) g of H by multiplying H 2 O mass by (2 / 18) Percent by Mass From formula: %mass of element = mass of element in 1 mole x 100 Molar mass of compound Dilutions Diluting a solution: Old molarity x old volume = new molarity x new volume...
View Full Document

Page1 / 2

chapter-03 - Have to “do experiment” to get actual...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online