Chapter 6 - Chemical Composition Flashcards
moles
- unit of measurement for atoms + molecules
6. 022 x 10^23
Avogadro’s No.
6.022 x 10^23
Molar Mass of a comound
the sum of each elements in a formula
ex) CO2
C: 12.01
O: 16.00 x 2 = 32.00
CO2: 12.01+32.00= 44.01g/mol
conversion of moles to atoms
6.022x10^23 atoms of element
/1 mole of element
ex) .5 mol He (6.022x20^23 atoms/1mol)
=3.011x10^23 atoms He
conversion of atoms to moles
1 mole of element
/6.022x10^23 atoms of element
ex) 3.011x10^23 atoms He (1 mol/6.022x10^23atoms)
= 0.5 mol He
Molar Mass
mass of 1 mol of an element (g/mol)
ex) Copper’s atomic mass is 63.55amu
1 Cu’s molar mass is 63.55g/mol
or
12.01g C= 1 mol C= 6.022x10^23 C atoms
Formula Unit Mass
the sum of the atomic masses in a chemical formula
ex) CO2 C: 12.01 O: 16.00 x2=32 CO2= 12.01+32.00=44.01amu (formula mass) Molar mass of CO2 = 44.01g/mol CO2
conversion of formula unit mass to moles
(element g/mol)x(1 mol/formula mass g)
ex) 22.5g CO2
22.5gx1 mol CO2/44.01g
=0.511 mol CO2
Mass Percent
mass percent composition of an element
-element’s percentage of total mass of the compound
Mass Percent
formula
mass of X in a sample of the compound
/mass of the sample of the compound
ex).358g of Cr reacts w O to form .523g metal oxide
.358g/.523g=.685=68.5% Cr in CrO
Mass Percent
chemical formula
mass of X in 1 mol compound
/mass of 1 mol of compound
Empirical Formula
-simplest whole number ratio of each type of atom in a compound.
ex) hydrogen peroxide molecular formula is H2O2.
the empirical formula is HO
Empirical Formula
calculation
find the mol of each element and then use that mole as the subscript. Make it a whole number.
Molecular formulas from empirical formulas
can only be calculated if the molar mass of the compound is known
MOLECULAR FORM=emp.form x N
N=molar mass/emp.form molar mass
ex) the emp.form CH2O and its molar mass is 180.2 g/mol
N=180.2/(12.01+2*1.01+16)=6
Molecular form=CH2O * 6
=C6H12O6
