Dimensional Analysis, Moles and Mass, and You PDF

Summary

This document introduces dimensional analysis, moles, and mass in chemistry. It includes practice problems and example calculations.

Full Transcript

Dimensional Analysis, Moles and Mass and you! Let’s “connect the dots” from our Intro to the Mole Activity to the calculations that we’ll be doing a lot of! Mole Practice Problems Do the following problems in your notes. Show your work and how the units cross cancel...

Dimensional Analysis, Moles and Mass and you! Let’s “connect the dots” from our Intro to the Mole Activity to the calculations that we’ll be doing a lot of! Mole Practice Problems Do the following problems in your notes. Show your work and how the units cross cancel EVERY TIME! How many atoms are in 2.50 mol of Al? 2.50 mol x ______ = How many moles contain 3.58 x 1023 formula units of ZnCl2? Lab Tie In After doing the Intro to the Mole Activity we know that a mole of any element has a mass equal to its atomic mass in grams. A mole of Boron has a mass of 10.8 g. Also, a mole of a compound has a mass equal to the sum of all its elements’ atomic masses. A mole of HCl = 36.46 g [1.01g (H) + 35.45g (Cl)]. The mass in grams of 1 mol of any substance is called its molar mass Why does this work? Side note, don’t write… Remember that each neutron and proton contributes one amu to an element’s atomic mass, so the total P + N = atomic mass (Carbon 12 is the standard, 6P + 6N) Average atomic mass (on the periodic table) for all elements is based on this. Average atomic mass is not always a whole number because there are different isotopes for each element. We did these calculations in Unit 1. ( Still a side note, don’t write…) Moles are also based on Carbon 12. Because we are humans and can’t see atoms or molecules to count them, the mole is based on the number of atoms in 12 g of Carbon 12. Someone (not Avogadro) at some point did the math (not literal counting) and that’s where 6.02 x 1023 came from! So what’s the point? 23 6.02 x 10 rep. particles = 1 mole and 1 mole = molar mass of that rep. particle We can use this information to make conversion factors for problems dealing with mass, representative particles or moles. Use these: 1 mole & molar mass 6.02 x 1023 rep part. 1 mole (remember, molar mass = atomic mass with grams for units) Let’s Practice… Example 1: What is the molar mass of KOH? 39.10g (K) + 16g (O) + 1.01g (H)= 56.11g KOH Notice that molar mass always has units. It’s g (which is really g/mol) Example 2: How many moles in 2.85 g of CaC2? Ca 40.08g 2.85 g x 1 mol = C x 2 24.02g 64.1g 64.10g 0.0445 mol Example 3: What is the mass of 1.32 moles of Ca? Notice that I always show my work when 1.32 mol x 40.08 g = calculating the molar 1 mol mass of a compound or 52.9 g formula unit. Let’s Kick it up a Notch… Example 4, 2 step problems: How many atoms are in 4.6 g of silver? 4.6 g x 1 mol x 6.02 x 1023 atoms = 107.87g 1 mol 2.6 x 1022 atoms Mole Practice (because it makes you mol perfect…) What is the mass in grams of 1.60 x 1024 atoms Mn? 1.60 x 1024 atoms Mn x 1 mol x 54.94 g = 6.02 x 1023 atoms 1 mol 146 g Mn

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