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A lesson on the mole-mass relationship in chemistry, covering objectives such as calculating the mass of a given number of moles of an element or compound, and vice versa. It includes examples and step-by-step calculations for converting between moles, mass, and the number of particles using molar mass and avogadro's number. The lesson also features problem-solving exercises to reinforce understanding of these concepts, making it a useful resource for students learning stoichiometry and chemical calculations. It also poses questions to promote critical thinking.
Typology: Schemes and Mind Maps
2024/2025
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MOLE-MASS RELATIONSHIP
Mole of a substance Molar mass (g) Avogadro’s number 1 mol Lithium 6.94 g 6.022 x 10²³ Li atoms 1 mol NH₃ 17.03 g 6.022 x 10²³ NH₃ molecule 1 mol MgBr₂ 184.11 g 6.022 x 10²³ MgBr₂ formula Unit
(a)Mole Mass (g) (b) Mass (g) Mole
Calculate (a) the mass in grams of 1.25 mol Li and (b) the number of moles of NH₃ in a 345 g sample of the gas. (Li - 6.94 g/mol, N- 14.0069 g/mol, H-1.008 g/mol) a. Mole Mass b. Mass Mole
Calculate (a) the mass in grams of 1.25 mol Li and (b) the number of moles of NH₃ in a 345 g sample of the gas. (Li - 6.94 g/mol, N- 14.0069 g/mol, H-1.008 g/mol) (a) Mole Mass
of Moles of substance x Molar mass of the substance (g)
1 mole 1.25 mol Li x 6.94 g Li = 1 mol Li
Calculate (a) the mass in grams of 1.25 mol Li and (b) the number of moles of NH₃ in a 345 g sample of the gas. (Li - 6.94 g/mol, N- 14.0069 g/mol, H-1.008 g/mol) (b) Mass Mole Mass (g) substance x 1 mole Molar mass of the substance (g) 345 g NH₃ x 1 mole NH₃ = 17.031 g NH₃
Calculate (a) the mass in grams of 1.25 mol Li and (b) the number of moles of NH₃ in a 345 g sample of the gas. (Li - 6.94 g/mol, N- 14.0069 g/mol, H-1.008 g/mol) (b) Mass Mole Mass (g) substance x 1 mole Molar mass of the substance (g) 345 g NH₃ x 1 mole NH₃ = 20.3 mol NH₃ 17.031 g NH₃
HOW DO YOU THINK CAN THE MOL AR
MASS BE REL ATED TO AVOGADRO’S
NUMBER?
HOW DO YOU THINK CAN THE MOL AR
MASS BE REL ATED TO AVOGADRO’S
NUMBER?
RP (Avogadro’s number) Mass Mass RP (Avogadro’s number)
How many atoms are in 11.3 g Pb?
Pb = 207.2 g
How many atoms are in 11.3 g Pb?
Mass RP
11.3 g Pb x 1 mole Pb x 6.022 x 10²³ Pb atoms = 3.28 x
10²² Pb atoms
207.2 g Pb 1 mole Pb
How many atoms are there in 44.2 g Mg? Mg = 24.3050 g 44.2 g Mg x 1 mole Mg x 6.022 x 10²³ Mg Atoms 24.3050 g Mg 1 mole Mg = 1.10 x 10²⁴ Mg atoms
Calculate the mass in grams of the following samples: a. 1.50 x 10¹⁸ atoms Ne Ne= 20.1797 g b. 8.35 x 10²¹ molecules CH₄ C= 12.0107, H=1.