Types of Reactions Essays - Chemical Properties, Mass, Free Essays

Types of Reactions Essays - Chemical Properties, Mass, Free Essays Types of Reactions -Synthesis A + B = AB -Decomposition AB = A + B -Single Displacement A + BC = AC + B -Double Displacement AB + CD = AD + CB -Combustion (metal burns with oxygen) C + O2 = CO2 Balancing Equations Must have equal number of atoms on each side of the equation based on the Law of Conservation of Mass. Steps to Balance an Equation -write out unbalanced equation -create a table with a column for reactants and products -balance equations using coefficients balance elements other than O and H balance polyatomic ions that appear unchanged balance other elements balance H and then O Isotopes Isotopes- atoms that have the same number of protons but different numbers of neutrons Radioisotopes- isotopes with unstable nuclei that decay into different often stable, isotopes Isotopic abundance- the amount of a given isotope of an element that exists in nature, expressed as a percentage of the total amount of this element Atomic Mass Unit (AMU) AMU- one twelfth of the mass of a carbon-12 atom Average Mass- sum of masses /# of masses Relative Atomic Mass (RAM) RAM- the atomic mass of an element in relation to that of another element To calculate RAM -RAM = m1f1 + m2f2 -Answer expressed in AMU (u) The Mole Mole- the SI based unit that is used to measure the amount of a substance; symbol is n; unit is mol Molar Mass- the mass of one mole of a substance; symbol is M; unit is g/mol Avogadros Constant- the number of particles in one mole of a substance; equal to 6.02 x 1023 particles Conversion Triangles Law of Conservation of Mass During a chemical reaction, the mass does not change, therefore the mass is conserved. Law of Definite Proportions Elements in a compound are always present in the same proportion by mass. e.g. water = 11.2% Hydrogen by mass 88.8% Oxygen by mass % mass of element=mass of element x100 mass of compound Mass percent- the mass of an element in a compound, expressed as a percentage of the total mass of the compound Percentage composition- the percent by mass of each element in a compound Method 1: Chemical formula known % composition of alanine (C3H7NO2) % mass of element=total mass of element molar mass M= (12.0x3)+(1.0x7)+(14.0)+(16.0x2) = 89.0 g/mol % carbon=12.0x3 x100 = 40.4% 89.0g/mol % hydrogen=1.0x7 x100 = 7.9% 89.0g/mol % nitrogen=14.0 x100 = 15.7% 89.0g/mol %oxygen=16.0x2 x100 = 36.0% 89.0g/mol Method 2: Chemical formula unknown A 24.5g sample of an unknown hydrocarbon is decomposed to yield 20.2g of pure carbon and 4.3g of hydrogen gas. Calculate % composition of this hydrocarbon. m of compound = 24.5g m of carbon = 20.2g m of hydrogen = 4.3g mass % C = mass of C x100 mass of compound = 20.2g x100 24.5g = 82.4% mass % of H= 100% - 82.4% = 17.6% Method 3: Composition from ratio When hydrogen combines with oxygen, it does so in a 1:8 ratio by mass. Calculate the % hydrogen. G- 1.0g H 8.0g O R- % composition A- % H = mass of H x100 mass of H+O M- % H = 1.0g x100 9.0g = 11% Empirical Formula Empirical formula- a formula that shows the smallest whole number ratio of the elements in a compound Determining Empirical Formula -Convert percentage composition data into mass data by assuming that the total mass of the sample is 100g -Determine the number of moles of each element in the sample by dividing the mass by the molar mass of each element -Convert the number of moles of each element into whole numbers that become subscripts in the empirical formula by dividing each amount in moles by the smallest amount -If the subscripts are not yet whole numbers, determine the least common multiple that will make the decimal values into whole numbers. Multiply all subscripts by this least common multiple. Use these numbers as subscripts to complete the empirical formula Empirical Formula from % Composition Calculate the empirical formula for a compound that is 85.6% carbon and 14.4% hydrogen G- % carbon 85.6% % hydrogen 14.4% assume 100g of C?H? R- empirical formula= ? nC=? nH=? A- nC= mC nH= mH MC MH M- nC= 85.6g 12.011g/mol nC=7.1268004 mol nH= 14.4g 1.00794 g/mol nH=14.286565 mol ratio of nC to nH = 7.1268004 mol 14.286565 mol ratio of nC to nH = 1:2 S- therefore the empirical formula is CH4 Molecular Formula Molecular formula- the formula for a compound that shows the number of atoms of each element that make up a molecule of that compound Determining Molecular Formula -Write the empirical formula -Determine the integer that