CH 101/Keys 2

From WolfWikis

Conversions in chemistry

A lot of conversions in chemistry center on:

1. from mass to moles and back using the molar mass in [g/mole] or Dalton
2. from moles to numbers of particle and back using Avogadro's number 6.022 10²³ [mole^-1]
3. from masses/moles/number of particles to concentrations in [mol/l] etc.
4. from volumes of gases at given P and T to moles, masses, numbers of molecules etc. (1 mole at P=1 atm and 25^oC = 22.4 liter)

Exercises

Exercise 1

Convert into moles (or [mmol] milimoles etc.) the following weights

1. 1.2 gram of B₂H₆

   M for this borane is 2*10.8+6*1=27.6 [g/mol]

   So putting it upside down we get: 1.2/27.6 [g] .[mol/g]= = 0.043478 [mol] = 43.5 [mmol]

2. 1.2 gram of B₂O₃

   M for this boron oxide is 2*10.8+3*16=69.6 [g/mol]

   1.2/69.6 [g] .[mol/g]= = 0.0172 [mol] = 17.2 [mmol]

3. 1.2 gram of B₂Te₃

   M for this boron telluride is 2*10.8+3*127.6= 404.4[g/mol]

   1.2/404.4 [g] .[mol/g]= = 0.002967 [mol] = 2.967 [mmol]

   (Notice how heavy it gets due tot the tellurium. No luxury to work in milimoles here!)

4. 2 mg of Ti₂Te₃

   M for this titanium telluride is 2*48+3*127.6= 478.8[g/mol]

   2mg / 478.8 [g] .[mol/g] = .00000417 mol = 4.17 μmoles.

5. 2 μg of SiCl₄

   M for silicon tetrachloride is 28+4*35.5= 170[g/mol]

   2/170 [μg] .[μmol/μg]= = 0.0117 [μmol] = 11.7 [nmol]

6. 63.546 g of Cu

   That's easy: 1 mole!

Exercise 2

The problem involves grams and moles: use the molar mass [g/mol] as conversion.

Must first compute it, then multiply by it to get rid of the moles.

Convert into grams (or kg or mg etc.)

1. 2 mmol of Sc₂Te₃

   M = 2* 45+ 3* 127.6 = 472.8 [g/mol] or [mg/mmol] or [μg/μmol] etc.

   2 * 472.8 [mmol]*[mg/mmol] = 945.6 [mg]

2. 2.34 μmol of Cs₂Te₃

   M= 2*133 + 3*127.6 = 648.8 [μg/μmol]

   2.34 * 648.8 [μmol][μg/μmol]= 1518.2 μg = 1.5282 mg

3. 2 mol of polyethylene with a molar mass of 120,000 Dalton

   2*120,000 = 240,000 [mol]*[g/mol]= 240 [kg]

   For a polymer 2 moles is a lot!!!

4. 0.234 mol of Sc₃LaGe₄O₁₄

   M= 3*45+139+4*72.6+14*16= 788.4

   0.234*788.4 [mol].[g/mol] = 184.5 [g]

Exercise 3

How many molecules or formula units are there in

1. 1.6 mol of acetone

   number of formula units (molecules) is 1.6*6.022 10²³ [mol][1/mol] = 9.9 10²³ ([dimensionless])

2. 1.7 nmol of Ag

   number of formula units (atoms) is 1.7*6.022 10²³ [nmol][1/mol] = 1.7*10 ^-9*10²³ = 10.5*10¹⁴ ([dimensionless])

   notice the extra factor of 10 ^-9 [mol/nmol].

3. 0.55 fmol of Si

   number of formula units (atoms) is 0.55 *6.022 10²³ [fmol][1/mol] = 0.55*10 ^-15*10²³ = 3.41*10⁸= ~ 34,100,000 ([dimensionless]) atoms

   Notice that even a femtomole still contains loads of atoms....

Exercise 4

How many molecules or formula units are there in

1. 12 fg of carbon-12

   The molar mass for carbon-12 is 12 [g/mol] by definition.

   We need to divide to get rid of grams and then multiply with N_Avo to get rid of the moles

   Finally we need to put in a factor of 10^-15 [g/fg] for the femto-prefix

   12/12*6.022 10²³*10^-15 [fg][mol/g][1/mol][g/fg] = 6.022.10⁸

2. 6 mg of Sc₂Te₃

   6 / 472.712 * 6.022 * 10²³ * 10^-3[mg][mol/g][1/mole][mg/g] = 7.87 X 10¹⁸.

3. 1 g of polyethylene with a molar mass of 120,000 Dalton

   Molar mass is 120,000 [g/mol]

   1/120,000*6.022 * 10²³ [g]*[mol/g]*[1/mol] = 5.16 10¹⁸

Exercise 5

How many atoms of oxygen are there in

1. 0.234 mol of Sc₃LaGe₄O₁₄

   number of moles of oxygen atoms 0.234*14 = 3.276 [mol]

   number of atoms 3.276* 6.022 10²³ [mol]*[1/mol] = 2.03*10²⁴

2. 2 mg of Ti₂Te₃

   zero!^[1]

3. 1.2 gram of B₂O₃

   17.2 [mmol] = 17.2 10^-3 [mol] ==> 17.2 10^-3 * 6.022 10²³ [mol]*[1/mol] = 1.06*10²²

Notes

1. ↑ Actually, one might wish... Often it is hard to keep oxygen out completely. There may be trace amounts present depending on how you synthesize the telluride