Chemistry Chapter 3 & 4 – Review – Flashcards

The molecular mass is the sum of the atomic masses of all the atoms in a molecule of the substance whereas the formula mass is the sum of the atomic masses of all the atoms in one formula unit of the compound, whether the compound is molecular or not. A given substance could have both a molecular mass and a formula mass if it existed as discrete molecules.

To obtain the formula mass of a substance, sum up the atomic masses of all atoms in the formula of the compound.

A mole of N2 contains Avogadro's number (6.02 x10²³) of N2 molecules and 2 x 6.02x10²³ atoms. One mole of Fe2(SO4)3 contains three moles of SO4²? ions, and it contains twelve moles of O atoms.

A sample of the compound of known mass is burned, and CO2 and H2O are obtained as products. Next, you relate the masses of CO2 and H2O to the masses of carbon and hydrogen. Then you calculate the mass percentages of C and H. You find the mass percentage of O by subtracting the mass percentage of C and H from 100.

The empirical formula is obtained from the percentage composition by assuming for the purposes of the calculation a sample of 100 g of the substance. Then the mass of each element in the sample equals the numerical value of the percentage. Convert the masses of the elements to moles of the elements using the atomic mass of each element. Divide the moles of each by the smallest number to obtain the smallest ratio of each atom. If necessary, find a whole-number factor to multiply these results by to obtain integers for the subscripts in the empirical formula.

The empirical formula is the formula of a substance written with the smallest integer (whole number) subscripts. Each of these subscripts in the formula C6H12O2 can be divided by 2, so the empirical formula of the compound is C3H6O.

The number of empirical formula units in a compound, n, equals the molecular mass divided by the empirical formula mass. n = 34.0 amu / 17.0 amu = 2.00 The molecular formula of hydrogen peroxide is therefore (HO)2 or H2O2

The coeffiecients in a chemical equation can be interpreted directly in terms of molecules or moles. For the mass interpretation, you will need the molar masses of CH4, O2, CO2, and H2O which are 16.0, 32.0, 44.0 and 18.0 g/mol. CH4 1 molecule 1 mole 16.0 g 2O2 2 molecules 2 moles 2 x 32.0 g

A chemical equation yields the mole ratio of a reactant to a second reactant or product. Once the mass of a reactant is converted to moles, this can be multiplied by the appropriate mole ratio to give the moles of a second reactant or product. Multiplying this number of moles by the appropriate molar mass gives mass. Thus, the masses of two different substances are related by a chemical equation.

The limiting reactant is the reactant that is entirely consumed when the reaction is complete. Because the reaction stops when the limiting reactant is used up, the moles of product are always determined by the starting number of moles of the limiting reactant.

Two examples are given in the book. The first involves making cheese sandwhiches. Each sandwhich requires two slices of bread and one slice of cheese. The limiting reactant is the cheese because some bread is left unused. The second example is assembling automobiles. Each auto requires one steering wheel, four tires, and other components. The limiting reactant is the tires, since they will run out first.

Since the theoretical yield represents the maximum amount of product that can be obtained by a reaction from given amounts of reactants under any conditions, in an actual experiment you can never obtain more than this amount.

a. 3.27 g NH3

b. 1 g of formaldehyde CH2O

d. 4.85 x10²?atoms

a. C3H4O3

Some electrolyte solutions are strongly conducting because they are almost completely ionized, and others are weakly conducting because they are weakly ionized. The former solutions will have many more ions to conduct electricity than will the latter solutions if both are present at the same concentrations.

A strong electrolyte is an electrolyte that exists in solution almost entirely as ions. An example is NaCl. When NaCl dissolves in water, it dissolves almost completely to give Na+ and Cl- ions. A weak electrolyte is an electrolyte that dissolves in water to give a relatively small percentage of ions. An example is NH3. When NH3 dissolves in water, it reacts very little with the water, so the level of NH3 is relatively high, and the level of the NH4+ and OH- is relatively low.

Soluble means the ability of a substance to dissolve in water. A compound is insoluble if it does not dissolve appreciably in water. An example of a soluble ionic compound is sodium chloride NaCl, and an example of an insoluble compound is calcium carbonate, CaCO3

The advantage of using a molecular equation to represent an ionic equation is that it states explicitly what chemical species have been added and what chemical species are obtained as products. It also makes stoichiometric calculations easy to perform. The disadvantages are: 1. the molecular equation does not represent the fact that the reaction actually involves ions. 2. the molecular equation does not indicate which species exist as ions and which exist as molecular solids or molecular gases.

A spectator ion is an ion that does not take part in the reaction. In the following ionic reaction, the Na+ and Cl- are spectator ions. Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ? Na+(aq) + Cl-(aq) + H2O(l)

A net ionic equation is an ionic equation from which spectator ions have been canceled. The value of such an equation is that it shows the reaction that actually occurs at the ionic level. An example is the ionic equation representing the reaction of calcium (CaCl2) with potassium carbonate (K2CO3). CaCl2(aq) + K2CO3(aq) ? CaCO3(s) + 2KCl(aq): Ca²?(aq) + CO3²?(aq) ?CaCO3(s) (net)

3 major reactions are: precipitation reactions, acid-base reactions, and oxidation-reduction reactions. (oxidation-reduction are: combination reactions, decomposition, displacement & combustion) -Precipitation reaction: involves the formation of an insoluble solid compound. 2KCl(aq) + Pb(NO3)2(aq) ? 2KNO3(aq) + Pb2(s) -Acid-base reaction (neutralization): results in an ionic compound and possibly water. HCl(aq) + NaOH(aq) ? NaCl(aq) + H2O(l) Oxidation-reduction: -combination reaction: 2 substances combine to form a third substance. 2Na(s) + Cl2(g) ? 2NaCl(s) -decomposition reaction: in which a single compound reacts to give 2 or more substances 2HgO(s) ? 2Hg(l) + O2(g) -displacement reaction: or single replacement: an element reacts w/ a compound displacing an element from it. Cu(s) + 2AgNO3(aq) ? 2Ag(s) + Cu(NO3)2(aq). -combustion reaction: reaction with oxygen, usually rapid release of heat to produce a flame. The products include one or more oxides. CH4(g) + 2O2(g) ? CO2(g) + 2H2O (l)

To prepare crystalline AgCl and NaNO3, first make solutions of AgNO3 and NaCl by weighing equivalent molar amounts of both solid compounds. Then mix the two solutions together, forming a precipitate of silver chloride and a solution of soluble sodium nitrate. Filter off the silver chloride and wash it with water to remove the sodium nitrate solution. Then allow it to dry to obtain pure crystalline silver chloride. Finally, take the filtrate containing the sodium nitrate and evaporate it, leaving pure crystalline sodium nitrate.

An example of a neutralization reaction is: HBr (acid) + KOH (Base) ? KBr (salt) + H2O(l)

An example of a polyprotic acid is carbonic acid, H2CO3. The successive neutralization is given by the following molecular equations: H2CO3 (aq) + NaOH(aq) ? NaHCO3(aq) + H2O(l) NaHCO3(aq) + NaOH(aq) ? Na2CO3(aq) + H2O(l)

Since an oxidation-reduction reaction is an electron transfer reaction, one substance must lose the electrons and be oxidized while another substance must gain electrons and be reduced.

A displacement reaction is an oxidation-reduction reaction in which a free element reacts with a compound, displacing an element from it. Cu(s) + 2AgNO3(aq) ? 2Ag(s) + Cu(NO3)2(aq) Ag+ is the oxidizing agent, and Cu is the reducing agent

The number of moles present does not change when the solution is diluted.

The reaction is: HCl + NaOH ? NaCl + H2O After titration, the volume of hydrochloric acid is converted to moles of HCl using the molarity. Since the stoichiometry of the reaction is 1 mol HCl to 1 mol NaOH, these quantities are equal. Moles HCl = moles NaOH = molarity x volume You could then multiply by the molar mass of NaOH to obtain the amount in the mixture.

e. HF(aq) + OH-(aq) ? F-(aq) + H2O(l)

c. Magnesium hydroxide(s)

d. MgCl2

d. HNO2

...

c. 200. g/mol

12

e. SrSO4

e. 0.004525

e. C4H4 Carbon = 92.3 /12(amu) Hydrogen = 7.7/1 (amu) C = 7.69, H= 7.7 To find molecular weight take compound weight 29.74 g and divide by the moles 0.572 moles = 51.99 g/mol MW = 52 EW = 13 (amu of both elements) MW/EW = 4 molecular formula is C4H4

d. H+(aq) + OH-(aq) ? H2O(l)

a. 21.2%

d. C5H14

e. Mg²?(aq) + CO3²?(aq) ? MgCO3(s)

a. HBr

I, II and IV only The mass of salt added, molar mass of salt and the total volume of the solution.

d. 1.1 L

a. NaCl

d. +2

a. removal of electrons

b. HC2H3O2(aq) + OH-(aq) ? C2H3O2-(aq) + H2O(l)