Problem 1: (2 pt.) Add up amu between the terms molecular mass and formula mass.
ID: 1030052 • Letter: P
Question
Problem 1: (2 pt.) Add up amu between the terms molecular mass and formula mass. Mass iA Problem 2: (3 pts.) What is the molar mass of the following elements? a. 1 mole of helium (gas)-L074-grams of helium (gas). b. 1 mole of su 254925 erams of s. c. 1 mole of oxygen (gas)-TG grams of oxygen (gas). a. How many molecules make up one mole of Sa?mot sa b. How many atoms of sulfur make up one mole of S8? Problem 3: (2 pts.) (Show all work. Apply the factor label metho)2zxio Problem 4: (1 pt.) (Show all work. Apply the factor label method.) How many atoms copper are in an old penny made of pure copper and weighing 2.15 g? Problem 5: (1 pt.) (Show all work. Apply the factor label method.) What is the mass of 6.022 x 1024 atoms of sodium? Problem 6: (1 pt.) (Apply Factor Label Method to solve.) How many mole of sodium correspond to 1.0x 101s atoms of sodium? Problem 7: (2 pts.) (Show all work.) Calculate the molecular weight of the following molecules. b. CHaNH2 (methyl amine) a. C2HsOH (ethanol) 1l PageExplanation / Answer
1.
Molecular mass : Mass of one molecule of the compound. It is useed interms of covlent molecules.
Formula mass : It is useful in the case of ionic compounds.
2.
(a) 1 mol of He gas = 4.0 g. of He gas
(b) 1 mol of S8 = 8 * 32 = 256 g. of S8
(c) 1 mol of O2 gas = 32 g. of O2 gas
3.
(a) 1 molecule of S8 is made from 8 atoms of S
1 mol of S8 is made from 6.022 * 1023 S8 molecules.
(b)
1 mol of S8 = 8 * 6.022 * 1023 = 4.82 * 1024 atoms of S
4.
Moles of Cu = mass / molar mass = 2.15 / 63.5 = 0.0338 mol
Atoms of Cu in 2.15 g. of Cu = 0.0338 * 6.022 * 1023 = 2.04 * 1022 atoms of Cu
5.
Mass of 6.022 * 1023 Na atoms = 23 g.
Then,
Mass of 6.022 * 1024 Na atoms = 6.022 * 1024 * 23 / (6.022 * 1023) = 230 g.
6.
6.022 * 1023 atoms of Na = 1 mol of Na
Then,
1.0 * 1015 atoms of Na = 1 * 1.0 * 1015 / (6.022 * 1023) = 1.66 * 10-9 mol of Na
7.
(a)
Molecular weight of C2H5OH=2(at.wt.of C)+6(at.wt.of H)+1(at.wt.of O) = 2(12)+6(1)+1(16) = 46 amu
(b)
Moleucalr weight of CH3NH2=1(at.wt.ofC)+5(at.wt.ofH)+1(at.wt.ofN) = 1(12)+5(1)+1(14) = 31 amu