1. Solve for the MM and % of each element.
CH3COOH
NH4NO3
C11H22O11
2. what would be the mass of 15.00×10^21 atoms of Cobalt?
3.The empirical formula of a compound is SiH3. If 0.0275 mol of compund has a mass of 1.71g, what is the compound's molecular formula?
4.What's the persentage of water in CuSO4·5H2O?
5. A compound contains 58.5% C, 7.3%H and 34.1% N. What is the formula of the compound?
2010年11月30日星期二
Percentage Composition --Nov.29
Definition: Percentage by mass of “species” in a chemical formula.
Ex. What is the percentage composition of CO2?
*assume you have 1 mole
Total MM=44.0g/mol
Percentage of C=(12g/mol)/(44g/mol)×100% = 27.3%
Percentage of O=(16g/mol×2)/(44g/mol)×100% = 72.7%
*(27.3%+72.3%=100%)
Percentage composition tells you which elements are in the compound and how much of each there is.
Fomula(useless):
Mass of element
Percentage composition=—————————— ×100%
Mass of compound
2010年11月24日星期三
Mole Conversions
Mole Conversions
1. Coversion from Particles Particles↔mass ( 1 moles= 6.022×10^23 particles)
a)particles →moles
b)moles→particles/molecules/formula units atoms
2.Concersions between moles↔molecules
a)moles→grams
b) grams→ mole
Harder Mole Conversions
Grams↔Moles↔Particle↔# Atom in a Practice
a)Conversions between particles to mass.
Ex. What is the mass of 2.78×10^22 Fe atoms?
2.78×10^22 ÷1mole/6.022×10^23 molecules × 80.1g/1mole=1.1×10^-3
b)Conversions between mass to particles.
How many atoms of Iron in 20.0g of Iron?
20.0g×1mole/55.8g×6.022×10^23atoms/1 mole=2×10^23
1. Coversion from Particles Particles↔mass ( 1 moles= 6.022×10^23 particles)
a)particles →moles
b)moles→particles/molecules/formula units atoms
2.Concersions between moles↔molecules
a)moles→grams
b) grams→ mole
Harder Mole Conversions
Grams↔Moles↔Particle↔# Atom in a Practice
a)Conversions between particles to mass.
Ex. What is the mass of 2.78×10^22 Fe atoms?
2.78×10^22 ÷1mole/6.022×10^23 molecules × 80.1g/1mole=1.1×10^-3
b)Conversions between mass to particles.
How many atoms of Iron in 20.0g of Iron?
20.0g×1mole/55.8g×6.022×10^23atoms/1 mole=2×10^23
2010年11月17日星期三
The Mole
-equal volume of different gases has a constant ratio
ex oxygen:hydrogen=16:1
carbon dioxide:hydrogen=22:1
-expressed by comparing it to the mass of another object
Avogadro's Hypothesis
Equal volume of different gases at the same temperature and pressure have the same number of particles.
If they have the same number of particles, the mass ratio is due to the mass of the particles.
Atomic Mass
Mass of 1 atom of the element in atomic mass units(amu or u or Dalton)
ex fluorine=19.0 amu
Formula Mass
All the atoms of a formula of an ionic compound(in amu)
ex potassium fluoride (KF) 39.1+19.0=581amu
Molecular Mass
All the atoms of a formula in a covalent compound (in amu)
ex carbon dioxide(CO2) = 12.0+16.0×2=44.0 amu
Molar Mass
Atomic/Molecular/Formula mass of any pure substance(in grams per mole)
ex 1mole of oxygen=16.0 g/mol
1mole of carbon =12.0 g/mol
(same number of particles)
Avogadro's Number
The number of particles in 1 mole of any amount of substance is
6.022×10^23 particles/mol
The mole is extremely important to chemists because it allows them to count atoms and molecules.
ex oxygen:hydrogen=16:1
carbon dioxide:hydrogen=22:1
-expressed by comparing it to the mass of another object
Avogadro's Hypothesis
Equal volume of different gases at the same temperature and pressure have the same number of particles.
If they have the same number of particles, the mass ratio is due to the mass of the particles.
Atomic Mass
Mass of 1 atom of the element in atomic mass units(amu or u or Dalton)
ex fluorine=19.0 amu
Formula Mass
All the atoms of a formula of an ionic compound(in amu)
ex potassium fluoride (KF) 39.1+19.0=581amu
Molecular Mass
All the atoms of a formula in a covalent compound (in amu)
ex carbon dioxide(CO2) = 12.0+16.0×2=44.0 amu
Molar Mass
Atomic/Molecular/Formula mass of any pure substance(in grams per mole)
ex 1mole of oxygen=16.0 g/mol
1mole of carbon =12.0 g/mol
(same number of particles)
Avogadro's Number
The number of particles in 1 mole of any amount of substance is
6.022×10^23 particles/mol
The mole is extremely important to chemists because it allows them to count atoms and molecules.
2010年11月3日星期三
Density Nov.11
1cm^3 of watere(H2O) = 1mL
Density of water = 1.0g/mL
or =1000g/L
For a solid →g/cm^3
For a liquid →g/mL
V=M/D
D object > D liquid sink
D object < D liquid float
Density of water = 1.0g/mL
or =1000g/L
For a solid →g/cm^3
For a liquid →g/mL
V=M/D
D object > D liquid sink
D object < D liquid float
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