Ch.11 Chemistry Mole – Flashcards

the SI Base unit used to measure the amount of a substance. It is the number of representative particles, carbon atoms, in exactly 12 g of pure carbon 12 We use the term the Mole to represent Avogadro's number A mole of atoms is 6.022 x 10 23 atoms, much like a dozen atoms is 12 atoms. AMU of Aluminum = 26.98 AMU of Copper is =63.55 AMU of Carbon is 12.01 So, 26.98 grams of aluminum, 63.55 grams of copper, and 12.01 grams of carbon each contain 6.022 x 10 23 atoms! 26.98 g of Al = 1 mole Al = 6.022 x 1023 atoms

6.02 x 10 to the 23 in 1811 Avogadro determined the volume of one mole of a gas. 3.5 mole sucrose x (6.02 x10 23 molecules of sucrose)/1 mol sucrose = 2.11 x10 24 molecules sucrose. There are 2.11 x 10 24 molecules of sucrose in 3.5 moles Number of moles x Av.Number = # of molecules

The mass in grams of one mole of any substance. The molar mass of any element is numerical equal to its atomic mass and has the units g/mol. Moles x molar mass = grams

Measuring out the average atomic mass of an element in grams is measuring out a mole of atoms or 6.022 x 10 23 atoms.

By adding the mass of each element in a compound.

10g Al x ( 1 Mol Al/26.98 g Al) = .371 Mol Al

10g Al x(1 Mol Al/26.98 g Al) x (6.011 x 10 23 Atoms Al)/1 Mol Al 1 Mol/ 6.022 x10 23 Atoms Atomic Mass of element in Grams/ 1 Mol Element

Al= 26.98 O=16.0 x3= 48 H= 1.01 x3 = 3.03 Molar mass = 78.01 Al = 26.98 is what % of 78.01? Al= 34.6%

Example: A sulfur compound contains 50% Sulfur and 50% O Find the Empirical Formula 100 g of compound contains 50g S and 50 g of O No of Moles of Atoms 50/32.07 = 1.56 mol S 50/16.0 = 3.114 mol of O Empirical formula = SO2 There is twice as many moles of O as there is S

A compound is found to be made up of 71.9% Carbon, 12.13% Hydrogen and 15.97% Oxygen The Compound has a Molecular weight (Molar Mass) of 300.54 g Determine the actual chemical formula for the compound. 100 grams of compound contains 71.9g C, 12.13g H, 15.97g O No of Moles of Atoms 71.9g C / 12.01 = 5.9 9 Moles of C 12.13g H /1.01 = 12 Moles of H 15.97 g O / 16.0 = 1 Mol of O Empirical Formula = c6h12o Molar Mass of Empirical Formula =100.18 Molecular Formula = C18H36O3

An analysis of a hydrate is found to contain 639% Copper II Sulfate (CuSO4) and 36.1% Water Calculate the simplest formula for this hydrate. 100g compound contains 63.9% CuSO4, 36.1% H2O # of Moles of Atoms 63.9/159.62 = .4 Moles of CuSO4 2.003/0.400 = 2.003 moles of H2O Empirical Formul o the Hydrate = CuSO4 ¥ 5H2O