Question

ercent Composition, Empirical and Section 3 - Calculations (Moles, Percent Composition, Emp Molecular Formulas) In order to earn full credit, you must show your work clearly, include units significant figures 1. (pts) You are given 3.50g of TNI, CHN Tou are given 3.50g of TNT. C-HSN:0. Use this starting amount to determine the quantities requested below. k clearly, include units and use reasonable a) Determine the molar mass of TNT: g/mol (no work necessary for this one). b) Determine the number of moles of TNT present in the sample. c) Determine the number of molecules of TNT present in the sample. 2. (6pts) The density of chloroform, CHCl3, is 1.50g/mL. How many molecules of chloroform are present in a 35.0-ml sample?

Answer 1

V NOZ 02 0 TNT, C/ H5N₂O6 Compound - Trinitro towene as we have to calculate the Molar mass of TNT, → Molar mass of carbon (C) = 12.011 g. moi Molar mass of Hydrogench) = 1.01 g. mort Molar 1 1 Nitrogen (N = 14.01g.moly Molar mass of oxygen (0) = Now, molar mass of TNT EM M = [7x12:01) + (5X101] + (3X14:01] +(6x1599) = 227.10 g.more Molar mass of TNT = 227-10 g.molt b) No. of moles! = Amount of comp(ö)/Molar mass (amory 227.10 g of INT 3 I mole 3.50 g of TNT 3.50 moles 227110 15.990 g.mov = 0.015 moles O No of moles of TNT present in the sample = 0.015 moles es No. of molecule: I mole corresponds to the mass at substance that contains 6-023 x 10 23 particles of the substance

Imole sample contains 6-023 X1023 no. of atoms 0.015 moles 1 0.015 X6023X1023 no. et atoms 3 9.03 x10 21 no. of atoms/ - molecules * No. of molecules of INT present in sample = 9.03 X1021 y Density of chloroform, Chay = 1150 g/mL (Given) We know that, density = Volume mi) Mass of 35 ml sample = density x volume $150 g/m2. X 35 m2 -525 g : Molar mass of Chelg = 119.52 g-moln Ll1x12.011) + (1x101) + (3x 35-5]]g.moit . 3119.52 g molt . No- of moles at sample = 52:58 119.52 gimo 0.44 moles A more contain 6-023X1023 no. of molecules 0.44 mole " 0.44 X 6.023x1023 no. of molecule = 2.65 x1023 no. of molecule 2665 X0023 molecules present in 35.0 mL of CHCl3 sample o

© Glements: % comp : CH 57.79 8.50 12.011 1.01 33.71 14.01 Molar mas : No or moles: 57.79 8.50 12.011 1-01 33.71 14.01 24 - 4.8 Ć : H : N 4.8 : 89 : 2.4 => : 29 => 12 : 3.5 : 1) x2 3 4:7:2 Empirical formula = [(qHy Nzła Molar mass of the compount = 240-260 g-moll. a [ex12:01) + (7x101) + (2x4.01)] = 240-260 or, 83.13 2 = 240-260 CV 25 25.02. =3 . Molecular formula = [CAH2 N2) = C12 H₂1 NG Molar mass = (12x12.011 + 21X1-01 +6X19.01) 249.40 g.molt • Molecular formula = 299.90 gimont

© We have to prepare 0.250 Mj HCl, (2.501 Strength of commercial Hu = 12.5 m we know, vi si = V2 32 Itere, S, Strength of commercio Hel = 12.5M Vi = Volume of commerciou Hel S2 = Strength at He we have to prepared V2 = Volume of Hel prepared = 2.5L 3,= 12.5M, 52= 0.25 m 2 Ura? V2 = 2.5L 2.5 X 0.25 . U= U2 S2 = 20 12.5 I 0.05L = 0.05 X1000 ML - 50 mL otal volume of prepared Hel = 2.5 L -25 Xv6oo me 2500 ML Volume of 12. 5M) Hel = 50ml Volume of wate - 2450 ML A To prepare 0.25 M Hel (2.5L) we have to add 50 mL Hal of 1265 ml strength to 2450 ml at water. 2. 5L of 0.25 M H = 2450 ml water + 50 mL of 12:5my Hey