Chemistry Cards – Flashcards
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Number of atoms in a formula
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formula of substance, subscript numbers are used to indicate the number of atoms/groups of atoms of each elemetn in formula
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Exception to number of atoms in formula
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you do not write the number one: H2O, there is one oxygen but the subscript one is not present
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How many atoms of each element in a formula unit of ammonium carbonate: process
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look at each element as an individual: NH4, CO3
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How many atoms of each element in a formula unit of ammonium carbonate: process: step 2
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look at charges of each: NH4^+, CO3^2-
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How many atoms of each element in a formula unit of ammonium carbonate: process: step 3
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cross the charges and write formula together: (NH4)2CO3
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How many atoms of each element in a formula unit of ammonium carbonate: process: step 4
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find out how many of each individual element: N=2, H=8, C=1, O=3
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Atomic mass
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average mass of all atoms of an element as they occur in nature
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Based on defining the mass of a carbon from 12 atom to
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12 u
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the average is weighted by______ _______ ________
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natural isotope distribution
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molecular mass (formula mass)
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the sum of each atomic mass of each atom in a molecule
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Formula mass example: Calcium Phosphate: step 1
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write out given: Ca, PO4
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Formula mass example: Calcium Phosphate: step 2
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write out charges: Ca^2+, PO4^3-
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Formula mass example: Calcium Phosphate: step 3
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cross charges: Ca3(PO4)2
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Formula mas example: Calcium Phosphate: step 4
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find out how many of each: Ca=3, P=2, 0=8
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Formula mass example: Calcium Phosphate: step 5
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find out mass of each element then multiply by how many there are: Ca= 3*40.08=120.24 P=2*30.97=61.94 O=8*16=128
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Formula mass example: Calcium Phosphate: step 6
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add masses together, 310.18 u
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avogadro's number/avogadro constant, NA
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the experimentally determined number of atoms in a mole, 6.022*10^23
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example: 1 mole of hydrogen atoms
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6.022*10^23 hydrogen atoms weighs 1.01 g H (you look at the mass to determine this, 1 mole equals 6.022*10^23)
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example: 1 mole of H2O
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6.022*10^23 water molecules= sum of all parts H= 2*1.01=2.02, O=16 (added together they equal 18.02)
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example: how many CO2 molecules are in 2 moles of CO2: step 1
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write out 1 mole= 6.022*10^23 (as a reminder)
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example: how many CO2 molecules are in 2 moles of CO2: step 2
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multiply (6.022*10^23)*2 = 1.2 *10^24
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Molar mass
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in grams of 1 mole of a substance g/mol, a bridge between macroscopic quantity of matter, grams a particulate level quantity of matter, moles.
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Difficult: how many H atoms are in 1.0 kg of ammonia. step 1
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path: kg --> g --> mol --> molecules --> atoms
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Difficult: how many H atoms are in 1.0 kg of ammonia. step 2
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1.0 kg = 1000g/1kg =6.022*10^23molecules/17.04g =3H atoms/ 1 molecule ANSWER: 1.06*10^26
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Percent Composition
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%A = parts of A / total parts *100
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Percent Composition definition
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percent composition of a compound = percentage by mass of each element in the compound
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Example of percent composition: there are 12 C, 22 H, 11 O find percent of C. Step 1
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multiply mass of each by how many of the element. 12*12.01=144.12 (C), 22*1.01=22.22 (H), 11*16=176 (O)
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Example of percent composition: there are 12 C, 22 H, 11 O find percent of C. Step 2
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add the masses together: 144.12+22.22+176=342.34
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Example of percent composition: there are 12 C, 22 H, 11 O find percent of C. Step 3
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Single out Carbon: 12 carbons=144.12
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Example of percent composition: there are 12 C, 22 H, 11 O find percent of C. Step 4
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divide amount of carbon by total: 144.12/342.34=.4210
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Example of percent composition: there are 12 C, 22 H, 11 O find percent of C. Step 5
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multiply answer by 100 and you receive the percent of carbon: 42.10%
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Empirical Formula
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the lowest whole number ratio of atoms of the elements in a compound
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empirical formula example: C2H4
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reduces to CH2
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Empirical formula = Molecular formula
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(EF)n,
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Refer to previous example (EF=MF) C2H4
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Empirical formula = CH2 CnH2n Molecular formula = (CH2)2
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How to find empirical formula: step 1
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find masses of different elements in a sample of the compound
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How to find empirical formula: step 2
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convert masses into moles (by dividing by molar mass)
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How to find empirical formula: step 3
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determine ratios of moles
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How to find empirical formula: step 4
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express moles of the atoms as smallest possible ratio integers
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How to find empirical formula: step 5
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write the empirical form using the number for each atom in the integer ratio as subscripts
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Example: empirical form of a compound that analyzes as 79.95% C, 9.40% H, 10.65%O. Step 1:
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g --> moles --> molar ratio --> emp formula
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Example: empirical form of a compound that analyzes as 79.95% C, 9.40% H, 10.65%O. Step 2
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C= 79.95 1 mol/12.01 g = 6.66 H= 9.40 1 mol/1.01g =9.31 O= 10.65 1 mol/16 = .66
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Example: empirical form of a compound that analyzes as 79.95% C, 9.40% H, 10.65%O. Step 3
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Divide each by the smallest number 6.66/ .66=10.0 9.31/ .66=14 .66/.66=1
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Example: empirical form of a compound that analyzes as 79.95% C, 9.40% H, 10.65%O. HINT
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.5 MULTIPLY BY 2 .3 BY 3
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Example: empirical form of a compound that analyzes as 79.95% C, 9.40% H, 10.65%O. FINAL PRODUCT
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C10H14O
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the molecular form of a compound can be found by determination of the number of the ______________
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empirical formula
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the molecular form of a compound can be found by determination of the number of the _____________ EXAMPLE
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CH2--> C2H4, C3H6 etc (anything equivalent to the ratio 1/2)
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n= --------------------------
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n = molar mass of compound/molecular formula ------------------------------------------------------------- molar mass of empirical formula
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N= ------------------------- EXAMPLE: c5h10o and a molar mass of 258. Solution:
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divide 258/empirical formula (86.15g) =2.99 (round to 3). n=3
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N= ------------------------- EXAMPLE: C5H10O and a molar mass of 258. Final product/step
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multiply each subscript by 3 C5H10O --> C15H30O3
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On test we will need to determine empirical formula: be able to preform this process. number 1
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1. determine empirical formula
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On test we will need to determine empirical formula: be able to preform this process. number 2
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2. calculate molar mass of empirical formula
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On test we will need to determine empirical formula: be able to preform this process. number 3
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3. determine molar mass
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On test we will need to determine empirical formula: be able to preform this process. number 4
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4. divide masses to get n
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On test we will need to determine empirical formula: be able to preform this process. number 5
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5. write molecular formula
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stoichiometry
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quantitative relationships among substances involved in a chemical reaction established by the balanced equation for the reaction a stoichiometry problem asks "how much or how many?"
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conversion factors
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the coefficient in a chem equation gives us the conversion factor to convert from number of particles of 1 substance, grouped in moles, to the number of particles of any other substance
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Stoichiometry: steps, #1
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balance the equation
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stoichiometry steps #2
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follow this structure: gA-->molA --> molB--gB
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example of structure gA-->molA --> molB--gB combustion reaction: CH4+2O2-->CO2+2H2O
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10 g CH4 1mol CH4/16.05g CH4 1 mol CO2/1molCH4 44.01 gCO2/1 mol CO2 Multiply numerators then divide by denominators =27.4 g CO2
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the number we achieve by preforming stoichiometry is the theoretical yield
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in theory what we will get (the ideal number)
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Don't forget sig figs!
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i.e. 60.77, would normally round to 61 but we are looking for 1 sig fig (found in problem) so we would round down because 0 at the end of 60 does not count as a sig fig
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percent yield
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the experimental/actual yield of a chemical reaction is usually less than the theoretical yield; predicted by a stoichiometry calculation because: reactants may be impure, reaction may not go to completion, other reactions may occur, actual yield is experimentally determined
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percent yield formula
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% = actual yield/theoretical yield *100
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example of percent yield formula: determine the percent yield if 6.97 g of ammonia is produced from the reaction of 6.22 g of nitrogen with excess hydrogen
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6.22 gN2 1mol N2/28.02 gN2 2NH3 mol/1 mol N2 17.04 gNH3/1Mol NH3 =7.56 NH3 theoretical 6.97 (experimental)/7.56 gNH3(theoretical) times 100 = 92.2% yield
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Limiting reactants (VERY IMPORTANT)
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limits the amount of products being produced (1 ingredient runs out)
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Example of limiting reactants:
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C (g) + O2 (G) --> CO2 (G) There will be one extra carbon
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Excess reactant:
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the reactant initially present in excess (some will remain unreacted) (carbon is excess in the example).
