Chem 130 Ch3 – Flashcards

the sum of the atomic masses of all the atoms in a molecule of the substance. Ex. Water is 2 +16=18 amu

the sum of the atomic masses of all the atoms in a formula unit of the compound salt(NaCl) is 23+35.5=58.5 amu

the quantity of a given substance that contains as many molecules or formula units as the number of atoms in 12g of carbon-12(6.02E23)

6.02E23 Ex. a mole of ethanol equals 6.02E23 ethanol molecules

the mass of one mole of the substance. for all substances, the molar mass in g/mol is numerically equal to the formula mass in amu. Ex. the molar mass of C2H6O is 46 g/mol (because the amu is (12x2)+(1x6) +16= 46 amu)

molar mass in g/mol Ex. How many grams in .0654 moles of ZnI2? .0654mol(319 g ZnI2/1 mol ZnI2)=20.9 g ZnI2

the MASS percentages of each element in the compound. Ex. 40% C 60% H

mass% of element A=(mass of element A in the whole/ mass of the whole) x 100%

Convert the masses to moles, then divide each by the smaller one to get whole integers. If you get 1.25, multiply both by 4 etc Ex. .483g N(1 mol N/14 g N)= .0345 mol N 1.104 g O (1 mol O/16 g O)= .069 mol O .345/.345= 1 N .069/.345= 2O NO2

First, you assume you have 100 g of the substance, meaning that the percents will easily become grams. Then you follow the process as if they were masses Ex. 17.5%, so you assume 100g, so you'd have 17.5 g. Then convert all of the grams to moles then divide by the smallest etc

First, find the empirical formula. Then find that empirical formulas formula mass(by adding up amu's). Then, n=molecular mass/empirical mass. This n is the multiple to find your molecular mass. Ex. You find an empirical formula to be CH2O. You then find its empirical mass to be 30 amu (using amu's on the periodic table). Then divide the molecular mass(given in the problem as 60 amu, it will be given) by the empirical mass to get n. In this case it is 60/30=2. Then multiply this number to the empirical formula, so (CH2O)2= C2H4O2

First find moles of C and H, by making the grams of CO2 and H2O moles(by multyplying by the molar mass), then multiplying the mole to mole ratio (ex H2O to H would be (2/1) because there is 2 moles of H for every mole of H2O). Then, convert both moles of C and H to grams(using molar mass), then subtract these grams from the 10.51 sample to get grams of ) in the original sample. Then convert O grams to moles(molar mass), then divide the 3 moles by the smallest to get the integers. The answer for this question is C2H4O. To the molecular formula, divide 88.11/empirical formula mass, making the final answer c4h8o2

The calculation of the quantities of reactants and products involved in a chemical reaction. g A x (mol A/ g A) x (mol B/ mol A) x (g B/ mol B)= g B Ex KHSO4 (aq) + KOH (aq) K2SO4 (aq)+ H2O (l) how many grams of potassium hydroxide are required for the complete reaction of 30.5 grams of potassium hydrogen sulfate? 30.5 g KHSO4 (1 mol KHSO4/136.2 g KHSO4........1 mol KOH/1 mol KHSO4.........56.11 g KOH/molKOH) = 12.6 g KOH

The reactant that is entirely consumed when the reaction goes to completion. To find out which is the limiting reactant.. Calculate the amount of product that can be formed by the initial amount of each reactant. (A) The reactant that gives the smaller amount of product is the limiting reactant. (B) The smaller amount of product is the amount that will be formed when the limiting reactant is used up. Calculate the amount of the non-limiting reactant that is needed to use up the limiting reactant. Subtract the amount of non-limiting reactant needed to use up the limiting reactant from the original amount of non-limiting reactant. The difference is the excess amount of the non-limiting reactant.

actual yield/theoretical yield x 100% The actual yield is always less or equal to the theoretical yield.