Question

Help me solve and understand this:

NA = 6.022 x10 23 mole -! 1. Calculate the following: c) the number of moles contained in 4.239 of methane (CH4) (b) the number of atoms in 7.469 of Li © the number of molecules in 43.0 g of NHz (ammonia) d) the number of grams in 2.969 x 1022molecules of CCl4 let chloride) 2. Find the percent composition of each element in acetamide, CaH NO (please use two places past the decimal(ie. 50.00%). 3. Find the empirical formula of the following: О) Сіоні,No4 (b) Насснаснас наснасна E (C) F-si-si-F (2) C3F6 e (e) C16H 32 Oy 4.) Find the empirical formula for oxalic acid whose composition is 26.7%C, 2.20% H and 71.1%O by weight. 10 In a seperate experiment, the molecular Weight of oxalic acid was found to be approximately 90g/mol. Find the Molecular formula

Answer 1

1)

a)

Molar mass of CH4,

MM = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass(CH4)= 4.23 g

use:

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(4.23 g)/(16.04 g/mol)

= 0.2637 mol

Answer: 0.264 mol

b)

Molar mass of Li = 6.968 g/mol

mass(Li)= 7.46 g

use:

number of mol of Li,

n = mass of Li/molar mass of Li

=(7.46 g)/(6.968 g/mol)

= 1.071 mol

use:

number of atoms = number of moles * Avogadro’s number

= 1.071 * 6.022*10^23 atoms

= 6.447*10^23 atoms

Answer: 6.45*10^23 atoms

C)

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 43.0 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(43 g)/(17.03 g/mol)

= 2.524 mol

use:

number of molecules = number of mol * Avogadro’s number

number of molecules = 2.524 * 6.022*10^23 molecules

number of molecules = 1.52*10^24 molecules

Answer: 1.52*10^24 molecules

d)

use:

number of mol = number of molecules / Avogadro’s number

number of mol = 2.969*10^22 / 6.022*10^23 mol

number of mol = 4.93*10^-2 mol

Molar mass of CCl4,

MM = 1*MM(C) + 4*MM(Cl)

= 1*12.01 + 4*35.45

= 153.81 g/mol

use:

mass of CCl4,

m = number of mol * molar mass

= 4.93*10^-2 mol * 1.538*10^2 g/mol

= 7.583 g

Answer: 7.583 g

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