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The concept of molecular weights and molar masses of compounds. It provides examples of hydrogen fluoride (hf) and carbon tetrafluoride (cf4) to illustrate how molecular weights are calculated by adding the atomic weights of all the atoms in a compound. The document also discusses the relationship between moles of molecules and their weights, and provides problems and solutions for practice.
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A molecule is the smallest possible unit of a compound. It consists of two or more atoms chemically bonded together.
A molecule of hydrogen fluoride, HF, is made up of 1 hydrogen atom bonded to 1 fluorine atom. Since a hydrogen atom weighs 1.0 amu and a fluorine atom weighs 19.0 amu, 1 molecule of HF weighs 1.0 amu + 19.0 amu = 20.0 amu.
A molecule of carbon tetrafluoride, CF 4 , is made up of 1 carbon atom bonded to 4 fluorine atoms. Each carbon atom weighs 12.0 amu and each fluorine atom weighs 19.0 amu so 1 CF 4 molecule weighs 1×12.0 amu + 4×19.0 amu = 88.0 amu. This number, 88.0, is called the molecular weight (or molecular mass) of CF4. It is obtained by adding the atomic weights of all the atoms in the compound.
The molecular weight of a compound is the sum of all the atomic weights in a molecule. Just like atomic weights are the relative masses of atoms relative to carbon-12, which is assigned a mass of 12.00000, molecular weights are relative masses of molecules relative to the mass of a carbon- atom. For example, the molecular weight of C 2 H 5 Cl is 2×12.01 + 5×1.008 + 1×35.45 = 64..
One mole of molecules is 6.022x10^23 molecules. One mole of HF is 6.022x10^23 molecules of HF. Since each HF molecule (weighing 20.0 amu) is 20.0 times as heavy as a hydrogen atom ( weighing 1.0 amu), 1 mole of HF molecules must weigh 20.0 times as much as 1 mole of H atoms. Since 1 mole of H atoms weigh 1.0 g, then 1 mole HF molecules must weigh 20.0 g.
So just like for atoms: 1 mole H = 6.022x10^23 atoms H = 1.0 g H ( the atomic weight in grams) 1 mole HF = 6.022x10^23 molecules HF = 20.0 g HF ( the molecular weight in grams) Likewise 1 mole CF 4 = 6.022x10^23 molecules CF 4 = 88.0 g CF 4
Also 1 mole CF 4 = 6.022x10^23 molecules CF 4 = 6.022x10^23 atoms C + 4×6.022x10^23 atoms F, since each molecule of CF 4 contains 1 atom of C and 4 atoms of F. So 1 mole CF 4 = 1 mole C + 4 moles F = 12.0 g C + 4×19.0 g F = 88.0 g CF 4
Try the following problems:
Answers:
Selected Solutions:
3 molec CF 4 = 3 molec CF 4 x 4
1 atom C 1 molec CF
= 3 atoms C
3.00 mol CF 4 = 3.00 mol CF 4 4
4 mol F 19.0 g F 1 mol CF 1 mol F
(^) = 228.0 g F
3.00 g CF 4 = 3.00 g CF 4
23 4 4
6.022x10 molec CF 88.01 g CF
= 2.05x10^22 molec CF 4
3.00x10^22 molec SnF 4 = 3.00x10^22 molec SnF 4 23 4 4
194.7 g SnF 6.022x10 molec SnF
= 9.70 g SnF 4
= 3.00x10^22 molec SnF 4 23 4 4 4
1 mol SnF 4 mol F 6.022x10 molec SnF 1 mol SnF
^ = 0.199 mol F
23
7
1 mol I 6.022x10 atoms I 259.9 g IF 1 mol I
= 6.95x10^21 atoms I
3.00 g IF 7 = 3.00 g IF 7 7 7
1 mol IF 259.9 g IF
= 0.0155 mol IF 7
3.00 g IF 7 = 3.00 g IF 7 7
7 mol F 19.00 g F 259.9 g IF 1 mol F
= 1.54 g F