This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Ison CH101 Name:_________, ___________ last, first Chapter 1 Worksheet Key Grams of substance Moles of substance Number of particles Molar mass (g/mol) Avogadro’s # (6.022x10 23 particles / mol ) Grams of substance Moles of substance Number of particles Molar mass (g/mol) Avogadro’s # (6.022x10 23 particles / mol ) We will use the unit factor method for conversions. The key is to set up your equation so that the conversion factor cancels the unwanted units and you are left with the desired units. Example I: Q: How many moles of CaCO 3 are there in 5.36g CaCO 3 ? A: We are asked to convert mass to moles which means we need to use the molar mass of CaCO 3 in g/mol as the conversion factor. The unwanted units are grams (mass) and desired units are moles so we will multiply the mass by the reciprocal of the molar mass. So first we calculate the molar mass of CaCO 3 , and then set up the equation so that unwanted units cancel. M m (CaCO 3 ) = 100g/mol Practice I: Q: What is the mass of 0.31 mol CaCO 3 ? Hint: Now we are asked to convert mass to moles, opposite of the example above. Set up your conversion so that the unwanted units (mol) cancel. mol CaCO 3 g CaCO 3 g mol g CaCO mol . 31 1 100 31 . 3 = × 3 3 0536 . 100 1 36 . 5 CaCO mol g mol CaCO g = × Ison CH101 Name:_________, ___________ last, first Example II: Q: How many CaCO 3 particles are there in 12.6g of CaCO 3 ? A: Here we are asked to find the number of particles which means that we need to use Avogadro’s #. However, we can only use N A as a conversion factor when working with moles so we first need to convert grams of CaCO 3 to moles of the same....
View
Full Document
 Spring '08
 BIGHAM
 Chemistry, Mole, Stoichiometry, Chemical reaction, mol, mol CaCO3

Click to edit the document details