Chemistry Unit 5

Chemical Change

change in which something new is formed due to rearrangement of chemical bonds


Chemical Reaction

change that occurs due to combining, separating or rearranging atoms -indicators are: formation of a precipitate. Production of a gas, color change, absorption and release of heat


Type of Chemical Reactions:

1.Single Displacement

2.Double Displacement

3.Synthesis

4.Combustion

5.Decomposition

1.free element and compound as reactants and products ex.)Fe+CuSO4 = FeSO4+Cu (look at activity series to determine if they can element and compound can actually react)

2.two compounds on each side ex.)KI+Pb(NO3)2 = KNO3+PbI2

3.two or more elements make one product ex.)Mg+O2 = MgO

4.the reaction with O2 as reactant…typically forms carbon dioxide and water ex.)CH4+2O2 = CO2+2H2O 5.one reactant(a compound) that breaks into 2 or more products ex.)C12H22O11 = C+H2O


Law of Conservation of Mass

Mass can neither be created nor destroyed. We balance equations to show that atoms are neither created nor destroyed in a chemical reaction


Guidelines for Balancing Equations

Start with an unbalanced equation, commonly referred to as the “skeleton” equation. Make an “element inventory”(listing the present elements as reactants and as products and also listing the quantity of each – keep current throughout problem as you change coefficients to balance. Change ONLY coefficients(numbers in front of formulas) until the number of atoms of each element is the same on both sides. Coefficients must be in the lowest whole number.

*Tip:leave free elements for last when changing coefficients.


Common Acids

HCl- hydrochloric

HBr- hyrobromic

HF- Hydroflouric

H2SO4– sulfuric acid

H2SO3– sulfurous acid

H3PO4– phosphoric acid

HNO3– nitric acid

HNO2– nitrous acid

H2CO3– carbonic acid

HC2H3O2– acetic acid


Diatomic Elements

elements that form molecules that consist of any 2 of the same type of atoms: Hydrogen(H2) Nitrogen(N2) Oxygen(O2) Florine(F2) Chlorine(Cl2) Bromine(Br2) Iodine(I2)


Activity Series

a list of elements organized according to the ease with which the elements undergo certain chemical reactions. Metals: Greater activity=greater ease of loss of electrons Nonmetals: greater ease of the gain of electrons Determined by single displacement reactions. Most active element at the top – can replace anything below it in a single displacement reaction. Cannot replace anything above it.

ex.)Li+NaCl = LiCl+Na…lithium, which was alone as a reactant stole chlorine from sodium as the product because lithium is higher on the activity series than sodium is, thus more reactive


Mole

SI unit for amount of substance. The mass in grams of a mole is equal to the mass of the substance in a.m.u.(atom or compound)

ex.)1 mole of carbon = 12gC

2 mole helium = 8gHe

3 mole of nitrogen = 42gN


Molar Mass

the mass of 1 mole of a pure substance(includes elements and compounds). Units:g/mole The molar mass of an elements is numerically equal to the atomic mass of the element in atomic mass units(amu) ex.)molar mass of Li = 6.9g/mol molar mass of S = 32.1g/mol

*round molar mass of pure substance to 1 decimal place A molar mass of an element contains 1 mole of atoms

ex.)22.99g of N, and 9.01g of Be all contain 1 mole of atoms When you take the molar mass of a compound, take the mass of each element, multiply the mass by how many elements are present, then add all the individual masses together

ex.)Ca3(PO4)2 = 310.3g/mol


3 Important Equivalences

1. Molar Mass

2. Avagadro’s Number

3. STP

1. 1 mole= the gram of the pure substance; “molar mass” (Use anytime you see grams in your problem – starting with or solving for…207.2Pb = 1mole)

2. 1 mole = 6.02 x 1023 (for atoms, particles, ions or molecules)

3. 1 mole = 22.4 L…of any gas at Standard temperature and standard pressure (STP) – for volume


Percent Composition

percentage of the total of the compound contributed by each element(calculate to the “tenths” place)

Formula: [total atomic mass of element/total mass of the compound] x 100 = the % composition of the element in the compound


1.Molecular Formula

2.Empirical Formula

1.actual ration of elements in a compound

*only used with covalent bonds/compounds

2.simplest whole number ration of elements in a compound

*for ionic and covalent bonds


Determining the Empirical Formula

Start with the number of grams of each element given in the problem. If percentages are given, assume that the total mass is 100 grams so that the mass of each element is = to the percent given

ex.)60.4% of 100g = 60.4g

Convert the mass of each element to moles using the molar mass(have at least 4 decimal places). Divide each mole value by the smallest number of moles calculated and round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula. If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same factor to get the lowest whole number multiple.

i.e.)if one solution is 1.5, then multiply EACH solution in the problem by 2 to get 3. If one solution is 1.25, then multiply each solution in the problem by 4 to get 5.

*you cannot have a fraction of an atom!

 

Example)NutraSweet is 57.14%C, 6.16%H, 9.52%N, and 27.18%O. Calculate the empirical formula of NutraSweet.

– 57.14g x 1mole/12gC = 4.7617molC

– 6.16gH x 1mole/1gH = 6.16moleH

– 9.52gN x 1mole/14gN = .68molN

– 27.18gO x 1molr/16gO = 1.6988molO

Okay, the smallest mole value that was calculated was .68 moles of Nitrogen. Now divide each mole value by .68 and round to the nearest whole number:

4.7617/.68 = 7 molCarbon

6.16/.68 = 9 molHydrogen

.68/.68 = 1 molNitrogen

1.6988/.68 = 2.5 molOxygen

Since Oxygen was too far to round to a whole number, multiply ALL solutions by 2 to get the lowest whole number:

7 x 2 = 14 molCarbon

9 x 2 = 18 molHydrogen

1 x 2 = 2 molNitrogen

2.5 x 2 = 5 molOxygen

The answer is: C14H18N2O5!