Exam 3 Flashcards
understanding the molecular and mass information in a balanced equation
- balanced chm equation gives amnt of reactant and product molecules that participate in a chemical run
- coefficients in a balanced eq. give the relative # of molecules
- all atoms present in the reactants are accounted for in the products
- 1 molecule of ethanol reacts w/ 3 molecules of O2 to produce 2 molecules of CO2 and 3 molecules of H2O
- 1 mole of ethanol reacts w 3 moles O2 to produce 2 moles of CO2 and 3 moles of H2O
Cu2S(s) + Cu2O(s) →Cu(s) + SO2(g) (unbalanced)
After balancing the reaction, determine how many dozens of copper atomscan be produced from a dozen molecules of the sulfide and two dozen molecules of the oxide of copper(I). Also, how many molesof coppercan be produced from one mole of the sulfide and two moles of the oxide?
6 dozens and 6 moles
relationship between moles of reactants and moles of products
- balanced eq can predict moles of products that a given # of moles of reactants will yield
- 2 moles of H2O yield 2 moles of H2 and 1 mole of O2
- 4 moles of H2O yield 4 moles of H2 and 2 moles of O2
mole ratio
allows us to convert from moles of 1 substance in a balanced eq to moles of a second substance in the eq
coefficient of unknown / coefficient of known
Consider the following balanced equation:Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s)
How many moles of NaF will be produced if 3.50 moles of Na is reacted with excess Na2SiF6?
5.25 moles NaF
Propane, C3H8, is a common fuel used for heating in rural areas. Predict the number of moles of CO2 formed when 3.74 moles of propane is burned in excess oxygen.
oxygen.
C3H8+ 5O2→ 3CO2+ 4H2O
11.2 moles
steps for calculating the masses of reactants and products in chm rxns
- balance the eq for the rxn
- convert the masses of reactants/products to moles
- use balanced eq to set up appropriate mole
- use mole ratio(s) to calculate the # of moles of desired reactant or product
- convert moles back to mass
For the following unbalanced equation: Cr(s) + O2(g) → Cr2O3(s) How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen?
21.9 g Cr2O3
Tin(II) fluoride is added to some dental products to help prevent cavities. Tin(II) fluoride is prepared according to the following equation:
Sn(s) + 2HF(aq) SnF2(aq) + H2(g)
How many grams of tin(II) fluoride can be produced from 55.0 g of hydrogen fluoride if there is plenty of tin available to react?
215 g
Consider the following reaction:
PCl3+ 3H2O H3PO3+ 3HCl
What mass of H2O is needed to completely react with 20.0 g of PCl3?
7.87g H2O
Consider the following reaction where X represents an unknown element:
6X(s) + 2B2O3(s) B4X3(s) + XO2(g)
If 175 g of X reacts with diborontrioxide to produce 2.43 moles of B4X3, what element does X represent?
C
limiting reactant mixture
N2 (g) + 3 H2 (g) -> 2 NH3 (g)
- limiting reagent is the reactant that runs out 1st and limits the amnt of products formed
- H2 is limiting reactant for rxn
limiting reactants
- determine which reactant is limiting to calculate the amnt of products tht will be formed
- methane and water will react to form products according to the eq: CH4 + H2O -> 3 H2 + CO
a) H2O molecules are used up first, leaving 2 unreacted CH4 molecules - amnt of products that can form is limited by the water
- water is limiting reactant
- methane is in excess
steps for solving stoichiometry problems involving limiting reactants by solving for limiting reactant first (Approach A)
- write + balance eq for rxn
- convert known masses of reactants to moles
- using the #s of moles of reactants + the appropriate mole ratios, determine which reactant is limiting
- using the amnt of limiting reactant and appropriate mole ratios, compute the # of moles of desired product
- if required: convert from moles of product to grams of product, using the molar mass
example 3 Approach A:
You react 10.0 g of A with 10.0 g of B. Given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol, what will be the mass of the product produced?
A + 3B -> 2C
8.33 g C
when solving for the amount of excess left over we must find out how much was __ - how much we __ __
needed; started with
steps for solving stoichiometry problems involving limiting reactants by determining the # of moles of products first (Approach B)
- write + balance eq for rxn
- convert known masses of reactants to moles
- using approp mole ratios, compute # of moles of product formed if each reactant was consumed
a) choose least # of moles of product formed - if required: convert moles of product to grams of product, using the molar mass
Consider the equation: A + 3B 4C. If 3.0 moles of A is reacted with 6.0 moles of B, which of the following is true after the reaction is complete?
A is the leftover reactant because you only need 2 moles of A and have 3 moles
How many grams of NH3 can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process?
N2+ 3H2→ 2NH3
3.65 g NH3
percent yield
- important indicator of the efficiency of a particular lab/industrial rxn
- comparison of the actual yield of a product to its theoretical yield
% yield = actual yield / theoretical yield x 100%
theoretical yield
- the max amnt of a given product that can be formed when the limiting reactant is completely obtained
- the amnt of a product actually obtained
actual yield
the amnt actually produced, of a rxn, is usually less than the max. expected (theoretical yield)
You react 10.0 g of A with 10.0 g of B. Given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol, what will be the mass of the product produced?
A + 3B 2C
- We determined that 8.33 g C should be produced. 7.23 g of C is what was actually made in the lab. What is the percent yield of the reaction?
(7.23g C / 8.33g C) x 100% = 86.8%
find the percent yield of a product if 1.50g of SO3 is made from 1.00g of O2 + excess sulfur
2 S + 3 O2 -> 2 SO3
89.8%
find the mass of P4 needed to produce 85.0g PF3 with 64.9% yield
P4 + 6 F2 -> 4 PF3
46.1 g
when solving the molar mass of diatomic molecules, always
x 2
energy
the ability to do work and product heat
energy is needed to…
oppose natural attractive forces
law of conservation of energy
- energy can be converted from one form to another
- can neither be created nor destroyed
- total e content of the universe is constant
potential e
stored energy; due to position or composition
kinetic e
e of motion; due to motion of the object that depends on the mass of the object and its velocity
heat involves the transfer of e between 2 objects due to a __ __
temperature difference
work
force acting over distance
energy is a __ function
state
state function
- property tht doesn’t depend in any way on a system’s past or future
- changes independently of its pathway
work and heat are not __ __
state functions
change in elevation (is/isn’t) a state function
is
distance traveled from 1 location to another (is/isn’t) a state function
isn’t
temperature
measure of random motions of the components of a substance
a higher temp has more __ than a colder temp
energy
heat
flow of e between 2 objects due to a temp difference between the 2 objects
- the way in which thermal e is transferred from a hot object to a cold object
system
part of the universe on which we wish to focus attention
surroundings
include every this else in the universe
endothermic process
- heat flow is INTO a system
- ABSORBS e from surroundings
(+) q
exothermic process
- e flows OUT of the system
(-) q
energy gained by surroundings must be __ to the energy lost by the system
equal
feels warmer is (exothermic/endothermic)
exothermic
- giving off heat
feels colder is (exothermic/endothermic)
endothermic
- heat is trapped
is freezing water endothermic or exothermic?
exothermic
Endo or exo: your hand gets cold when u touch ice
exothermic
endo or exo: the ice gets warmer when u touch it
endo
endo or exo: water boils in a kettle being heated on a stove
endo
endo or exo: water vapor condenses on a cold pipe
exo
endo or exo: ice cream melts
endo
endo or exo: methane is burning in a Bunsen burner in a lab
endo
hydrogen and oxygen react violently to form water. Which is lower in energy, a mixture of hydrogen and oxygen, or water?
water
thermodynamics
- study of energy
- law of conservation of energy is often called the 1st law of thermodynamics
- e of universe is constant
the more ordered something is, the __ energy it has
less
internal energy (E)
- E of a system is the sum of the KE and PE of all the particles in the system
- ∆ in the E of a system is
∆E = q + w - sign reflects the system’s POV
- endothermic process: (+) q
- exothermic process: (-) q
internal energy formula
∆E = q + w
system does work on the surroundings, w is __
(-)
surroundings do work on the system, w is
(+)
energy units
- calorie
- joule
calorie
the amnt of e or heat required to raise the temp of 1 gram of water by 1ºC
1 calorie = __ joules
4.184
convert 60.1 cal to joules
251 J
the energy (heat) required to change the temp of a substance depends on
- amnt of a substance being heated (# of grams)
- temp change (# of degrees)
- identity of the substance (specific heat capacity)
specific heat capacity
the e required to ∆ the temp of 1 gram of a substance by 1ºC
formula to calculate energy required for a rxn
Q = mc∆T *Q= energy (heat) required (J) m= mass (g) c= specific heat capacity (J/g ºC) ∆T= change in temp (ºC) (Tfinal - Tinitial)
calculate the amnt of heat e (in joules) needed to raise the temp of 6.25g of water from 21.0ºC to 39.0ºC
*specific heat capacity of water: 4.184 J/g ºC
471 J
A 100.0 g sample of water at 90 °C is added to a 100.0 g sample of water at 10 °C
the final temp of the water is likely to be:
a) between 50ºC and 90ºC
b) 50ºC
c) between 10ºC and 50ºC
50ºC
A 100.0 g sample of water at 90.0 °C is added to a 500.0 g sample of water at 10.0 °C
1) The final temperature of the water is likely to be:
a) between 50ºC and 90ºC
b) 50ºC
c) between 10ºC and 50ºC
2) calculate the final temp of water
1) between 10ºC and 50ºC
2) 23.3ºC
You have a Styrofoam cup with 50.0 g of water at 10.0 C. You add a 50.0 g iron ball at 90.0 C to the water. (sH2O= 4.18 J/g °C and sFe= 0.45 J/g °C)
1) The final temperature of the water is likely to be:
a) between 50ºC and 90ºC
b) 50ºC
c) between 10ºC and 50ºC
2) calculate the final temp of the water.
1) between 10ºC and 50ºC
2) 18ºC
change in enthalpy (∆H)
- state function
- ∆H= q at constant pressure
- ∆Hp = heat
Consider the rxn:
S(s) + O2(g) → SO2(g)ΔH = –296 kJ per mole of SO2 formed
Calculate the quantity of heat released when 2.10 g of sulfur is burned in oxygen at constant pressure
-19.4 kJ
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)ΔH= –2221 kJ
Assume that all of the heat comes from the combustion of propane. Calculate ΔH when 5.00 g of propane is burned in excess oxygen at constant pressure
-252 kJ
Hess’s Law
- in going from a particular set of reactants to a particular set of products, the ∆H is the same whether the rxn takes lace in one step or in a series of steps
characteristics of enthalpy changes
- if a rxn is reversed, the sign of ∆H is also reversed
- magnitude of ∆H is directly proportional to the quantities of reactants and products in a rxn
- if the coefficients in a balanced rxn are multiplied by an integer, the value of ∆H is multiplied by that integer
1) NH3 (g) -> 1/2N2 (g) + 3/2H2 (g) ∆H= 46 kJ
2) 2 H2 (g) + O2 (g) -> 2 H2O (g) ∆H= -484 kJ
calculate the ∆H for the rxn:
2 N2 (g) + 6 H2O(g) -> 3 O2 (g) + 4 NH3 (g)
1268 kJ
Calculate ΔH for the reaction: SO2+ ½O2→SO3 Given: 1) S + O2→ SO2ΔH = –297 kJ 2) 2S + 3O2→2SO3ΔH = –792 kJ
-99 kJ
Hess’s Law: like terms
- same sides : ADD
- opposite sides: SUBTRACT
Given: 1) C(s) + O2(g) -> CO2(g) ∆H = -393 kJ 2) 2CO(g) + O2(g) -> 2CO2(g) ∆H = -566 kJ Calculate the ∆H for the rxn: C(s) + 1/2O2(g) ->CO(g)
-110 kJ
specific heat capacities of common substances
substance: specific heat capacity (J/g ºC):
water (l) 4.184
water (s) 2.03
water (g) 2.0
aluminum (s) 0.89
iron (s) 0.45
mercury (l) 0.14
carbon (s) 0.71
silver (s) 0.24
gold (s) 0.13
nuclear model of an atom
- atom has small dense nucleus tht is (+)
- nucleus has protons (+) and neutrons (0)
- remainder of atom is mostly empty space
- contains e- (-)
- nuclear charge (n+) is balanced by the presence of e- moving in some way around the nucleus
- atoms follow orders
nuclear charge is __ bc of __
balanced; e-
orbitals
- nothing like orbits!
- probability of finding an e- within a certain space around the nucleus was predicted
- the wave mechanical model gives no info abt when the e- occupies a certain pt in space/ how it moves
- don’t have sharp boundaries
- chemists define an orbital’s size as the sphere that contains 90% of the total e- probability
orbital size is defined as..
the sphere tht has 90% of total e- probability
hydrogen energy levels
- called principal energy levels
- they are indicated by whole #s
- what goes up must go down
- start at ground state and go up to n=4
e- configuration
how e- are arranged
-e- can jump from high to low e levels
what is the e- configuration of an H atom?
1s1
orbital diagram
orbital is denoted by a box grouped by sub level coordinating arrows to represent e-
e- configuration
s= 2 p= 6 d= 10 f= 14 *divide numbers above by 2 to determine the # of boxes for the orbital diagram
where can u find the number of e- for an atom?
periodic table
e- configuration of Li atom
1s1 2s1
e- configuration of an O atom
1s1 2s2 2p4
e- configuration of S
1s2 2s2 2p6 3s2 3p4
metals and non-metals
metals tend to lose e- to form (+) ions
nonmetals tend to gain e- to form (-) ions
ionization e
e required to remove an e- from a gaseous atom
*trends of the periodic table
X(g) -> X+(g) + e-
__ e- are bound much more tightly than __ e-
core; valence
as we go across a period from left to right, the ionization e ___
increases
as we go down a group from top to bottom, the ionization e ___
decreases
as we go across a period from left to right, the atomic radius __
decreases
as we go down a group from top to bottom, the atomic radius ___
increases
as we go down a group, the size of the atom __
increases
core e-
the inner e- that aren’t involved in binding atoms to each other
valence e-
e- in the outermost or the highest principal e level of an atom
- 1s2 2s2 2p6, where valence e- equal 8
- the elements in the same group have the same valence e- configuration
- elements w the same valence e- arrangement show very similar chm behavior
the elements in the same group have the same __
valence e- configuration
chemical bond
force that holds group of 2 or more atoms together and makes them function as a unit
bond e
e required to break a chem bond
ionic bonding
- metal + a non-metal
- e- are transferred from an atom that can lose e- relatively easily to an atom tht has a high affinity for e-
covalent bonding
aka non polar covalent
- equal sharing of e-
- e- are shared by nuclei
- same element
polar covalent bond
- unequal sharing of e- between 2 atoms
- one atom attracts the e- more than the other atom
- results in development of bond polarity (partial (+) and (-) charge)
electronegativity
relative ability of an atom in a molecule to attract shared e- to itself
- increases across a period and decreases down a group
- ranges from highest at 4.0 fluorine to lowest at 0.7 cesium and francium
as we go across a period, electronegativity __
increases
as we go down a group, electronegativity __
decreases
what is the most electronegative element?
fluorine
what is the least electronegative element?
cesium and francium
if lithium and fluorine react, which has more attraction for an e-?
fluorine
in a bond between fluorine and iodine, which has more attraction for an e-?
fluorine
rank the following from smallest to largest atomic radius:
Ar, S 2-, Ca 2+, K+, Cl-
Ca2+ < K+ < Ar < Cl- < S 2-
the cation is always __ than the anion
smaller
which atom or ion has the smallest radius?
a) O 2+
b) O +
c) O
d) O 2-
O 2+
the polarity of a bond depends on the difference between the __ values of the atoms forming in the bond
electronegativity
arrange the following bonds from most polar to least polar
a) N-F O-F C-F
b) C-F N-O Si-F
c) Cl-Cl B-Cl S-Cl
a) C-F, N-F, O-F
b) Si-F, C-F, N-O
c) B-Cl, S-Cl, Cl-Cl
which of the following bonds would be the least polar yet still be considered polar covalent?
Mg-O, C-O, O-O, Si-O, N-O
N-O
which of the following bonds would be the most polar w/o being considered ionic?
Mg-O, C-O, O-O, Si-O, N-O
Si-O
bond polarity: dipole moment
- property of a molecule whose charge distribution can be represented by a center of a positive charge and a center of (-) charge
- dipole character of a molecule is represented by an arrow aka vector
- arrow points to (-) charge center and its tail indicates the (+) center of charge
- tug of war
dipole moment in a water molecule
- polarity of water affects its properties
- permits ionic compounds to dissolve in it
- causes water to remain a liquid at the temp. on the earth’s surface
e- configurations of ions
- Representative (main-group) metals form ions by losing enough electrons to achieve the configuration of the previous noble gas
- Nonmetals form ions by gaining enough electrons to achieve the configuration of the next noble gas
e- configurations and bonding
- when a nonmetal and a Group 1, 2, or 3 metal react to form a binary ionic compound, the ions form in such a way that
a) the valence e- configuration of the nonmetal is completed to achieve the configuration of the next noble gas and
b) the valence orbitals of the metal are emptied to achieve the configuration of the previous noble gas - When two nonmetals react to form a covalent bond, they share e- in a way that completes the valence e- configurations of both atoms
com compounds are always electrically __
neutral
e- configurations tree
1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p
how to find the number of core and valence e-
look at the group # to determine the # of valence e- and subtract the atomic number by that to get the # of core e-
cations are always __ than their parent atoms
smaller
anions are always __ than their parent atoms
larger
lewis structures
- representation of molecules
- shows how valence e- are arranged among atoms in a molecule
- octet rule
- duet rule
writing lewis structures
- bonding pairs are shared between 2 atoms
- unshared pairs aka lone pairs are not shared and not involved in bonding
octet rule
outermost shell has 8 e-
steps for writing lewis structures
- sum the valence e- from all atoms
- use 1 pair of e- to form a bond between each pair of bound atoms
- arrange the remaining e- to satisfy the octet rule for each second-row element & duet rule for hydrogen
in lewis structures, everything must have __ rule
octet
exceptions to the octet rule
- boron can have 6 e-
- sulfur can have 10 e-
draw a lewis structure for BF3
24 e-
draw a lewis structure for CO2
.
draw a lewis structure for CCl4
w
draw a lewis structure for CN
.
linear structure
atoms arranged in a line
trigonal planar structure
atoms arranged in a triangle
diatomic have __ polarity
low
ionization energy
e required to remove an e- from a gaseous atom
electronegativity
relative ability of an atom in a molecule to attract shared e- to itself
dipole moment
property of a molecule whose charge distribution can be represented by a center of a positive charge and a center of (-) charge
dipole moments only occur in __ bonds
covalent
in dipole moments, the farther apart the atoms are, the more __
polar
find the molecular structure of NH3
tetrahedral
find the molecular structure of CH4
tetrahedral
as we go left to right, the atomic size __
decreases
how to find the amount of excess left over in rxn
1) moles of LR x mole ratio (moles of ER ÷ moles of LR) = moles used of ER
2) moles started with ER - moles used ER = moles ER left over
3) convert moles to grams by x molar mass of ER
exothermic has a __ w
(+)
endothermic has a __ w
(-)
system does work, w is __
(-)
surroundings do work, w is __
(+)
