Blue litmus paper turns ___ in acid. Phenolphthalein stays ____ in acid.
ammonia, sodium hydroxide (or lye) found in oven and drain cleaners, calcium hydroxide (lime), magnesium hydroxide (milk of magnesia), potassium hydroxide, sodium carbonate (washing soda), and calcium oxide (quicklime).
Red litmus paper turns ___ in base. Phenolphthalein turns ____in base.
Neutral: pH __: water.
*____: Any ionic compound that is a neutralization product of an acid-base reaction.
Which ion affects the pH and which one is the Spectator ion?
The Bunsen burner uses natural gas; a mixture of gaseous _______, but primarily _______.
Hydrocarbon methane, CH4
With an insufficient supply of oxygen, small carbon particles are produced, which, when heated to ______, produce a yello, luminous flame.
Identify the parts of the flame, nonluminous.
3 distinct cones.
Described as non-glowing; blue; produced as a result of there being a sufficient amount of oxygen; produces carbon dioxide & water as combustion products
Described as glowing; orange; due to lack of oxygen
Described as a molecule consisting of only the elements carbon and hydrogen
Define Intensive property
property independent of sample size
Objectives for Ex. 1: Basic Lab Operations
• To light and properly adjust the flame of a Bunsen burner
• To develop the skill for properly operating a balance
• To determine the density of an unknown substance
The solid is not completely submerged in water. Will this technique error increase, decrease or have no effect on the reported density?
this technique error will not change the mass of the solid, but will decrease the volume. When doing the formula d=m/v, this decrease in volume will result in an increase in density
Mass does not change but the volume will be incorrectly less.
What will this do to the formula M/V?
Volume is at the bottom so if it gets smaller, the density will increase
Suppose that after delivery several drops of the water cling to the inner wall of the pipet (bc the pipet wall is dirty). Will this technique error increase, decrease or have no effect on the reported density of water?
the volume would decrease. This decrease in volume would result in an overall increase in density.
The unknown liquid is volatile. If some of the liquid evaporates between the time the liquid is delivered to the beaker and the time its mass is measured, will the reported density of the liquid be too low, too high or unaffected?
both its volume and mass will decrease which will result in a lower density than was anticipated
Do all weighing papers have the same exact mass?
No, they are all slightly different.
The mass of a weighing paper must always be taken and tared.
A yellow luminous flame is produced when an insufficient amount of oxygen is supplied. This flame should not be used when heating because it will coat glassware with black soot. What element is black soot made of?
A graduated cylinder contains 15.75 mL of water. A metal that weighs 32.138 g is added.
The final reading is 38.55 mL. Using the correct number of significant figures, determine the density of the metal = _____ g/mL
1. First Determine Volume of Metal: 38.55-15.75=22.8
2. Second Determine Density of metal: D=M/V; 32.138/22.8
= 1.41 g/mL
Objective for Exp. 7: empirical formulas
• To determine the empirical formulas of one compound by combination reactions
• To determine the mole ratio of the decomposition products of a compound
Empirical formula & steps to determine it
the simplest whole-number ratio of moles of elements in the compound.
1. Determine the mass of each element in the sample.
2. Calculate the number of moles of each element in the sample.
3. Express the ratio of the moles of each element as small whole numbers.
two elements combine to form a compound
a compound decomposes into two or more elements or simpler compounds
a glass container or other apparatus holding a drying agent for removing moisture from specimens and protecting them from water vapor in the air
combination reaction of Magnesium and Oxygen: balanced equation
2 Mg (s) + O2 (g)--> 2MgO2(s)
Percent by Mass formula
part/whole x 100
Empirical formula of Magnesium oxide
The crucible is not fired, as the procedure suggests, but had retained some impurities from previous use (or it could be oily smudges from fingers). The mass of the "dirty" crucible is recorded. However the impurities are burned off in the experiment. Will the reported mass of the final product be too high, too low, or unchanged as a result of this technique error?
the reported mass of the final product will be too high
Javier forgot to polish the magnesium metal. Will the reported mole ratio of magnesium to oxygen be too high or too low as a result of his error?
the reported mole ratio of magnesium to oxygen will be too high. This is because not polishing the metal could mean that there is dirt on it making the mass higher than it is supposed to be, which could lead to thinking more magnesium reacted, ultimately giving a higher mole ratio of magnesium to oxygen.
In a hurry to complete the experiment, Sylvia did not allow all of the magnesium to react. Will her reported magnesium to oxygen ratio be reported too high or too low?
Sylvia's reported magnesium to oxygen ratio will be too high. No reaction means more magnesium which will ultimately cause a higher magnesium to oxygen ratio.
What is the correct order of steps for calculating the empirical formula below?
A sample of iron is chemically combined with chlorine gas to form iron chloride.
Determine the Empirical Formula from the following data:
Mass of crucible: 19.91 g
Mass of crucible and iron: 22.51 g
Mass of crucible and product: 27.46 g
Step 1: Find mass of iron and convert to moles by using the molar mass of iron.
How do you find the mass of iron?
Subtract (mass of crucible and iron) - (mass of crucible).
Step 2: Find mass of chlorine and convert to moles by using the molar mass of chlorine.
How do you find the mass of chlorine?
Subtract (mass of crucible and product) - (mass of crucible and iron).
Step 3: Divide all moles by the smallest number of moles.
You are trying to find the subscripts x and y for FexCly.
Step 4: Determine whole numbers for the subscripts:
If = .5, multiply by 2; if = .3, multiply by 3
it has reacted.
it has a +2 charge because it lost 2 electrons.
found in group II, everyone there will lose 2 electrons, memorize that.
it has not reacted.
it has no charge.
oxygen gas, you are breathing it now
it is found naturally as a diatomic molecule, held together by a covalent double bond.
it has no charge because it has not reacted with anything.
it has reacted.
it has a -2 charge because it gained 2 electrons.
found in group VI, everyone there will gain 2 electrons, memorize that
made from a combination reaction, where the 2 shall become one.
composed of 2 ions: Mg2+ and O2
At the end of the experiment, what is the point of heating and reheating the crucible and when should you stop?
Get mass after 1st heating and cooling, heat again for complete reaction and stop when you have achieved constant mass, the mass difference between this heating and the previous heating is less than 0.01 g. Take the last mass.
A sample of iron is chemically combined with chlorine gas to form iron(III) chloride.
What is the correct balanced equation?
2Fe(s) + 3Cl2(g) --> 2FeCl3(s)
Objective for Exp. 8: Limiting Reactant
• To determine the limiting reactant in a mixture of two soluble salts
• To determine the percent composition of CaCl2⋅2H2O and K2C2O4⋅H2O
Study of a chemical reaction using a balanced equation
The reactant determining the amount of product generated in a chemical reaction; runs out first
A chemical equation that presents ionic compounds in the form in which they exist in aqueous solution.
cations or anions that do not participate in any observable or detectable chemical reaction
an equation that includes only those ions that participate in the observed chemical reaction
Net ionic equation
If the step for digesting the precipitate were omitted, will the reported "percent limiting reactant" in the salt mixture be too high, too low, or unaffected?
The reported percent limiting reactant in the salt mixture would be too low. This would be due to the fact that the precipitate could possibly go through the filter and into the mixture affecting the limiting reactant, making it less than what it actually is
A couple of drops of water were accidentally placed on the properly folded filter paper before its mass was measured. However, in Part A.6, the CaC2O4•H2O precipitate and the filter paper were dry. As a result of this sloppy technique, will the mass of the limiting reactant be reported too high, too low, or remain unaffected?
As a result of this sloppy technique, the added water drops will add additional mass which will make the filter paper have a reportedly lower mass than it actually does
Because of the porosity of the filter paper some of the CaC2O4•H2O precipitate passes through the filter paper. Will the reported percent of the limiting reactant in the original salt mixture be reported too high or too low?
percent of the limiting reactant would be calculated too low. This is due to the possibility of a substantial amount of the precipitate going into the mixture affecting the limiting reactant, making it less than what it actually is
The CaC2O4•H2O precipitate is not completely air-dried when its mass is determined. Will the reported mass of the limiting reactant in the original salt mixture be reported too high or too low?
mass of the limiting reactant in the original salt mixture will be too high. This is due to the fact that the mass of the water will add more mass, making the overall mass higher than expected.
The drying oven, although thought (and assumed) to be set at 125°C, had an inside temperature of 84°C. How will this error affect the reported percent by mass of the limiting reactant in the salt mixture . . . too high, too low, or unaffected?
mass of the limiting reactant will be will be too high, this is because a lower temperature will result in the product not drying completely which in turn will mean that there is more product on the filter paper and more product means more mass
A reagent bottle on the shelf labeled 0.5 M NaCl was used in place of the 0.5 M CaCl2. Assuming C2O42- to be in excess, what would be observed as a result of using this wrong reagent in this test?
there would be no reaction since Na doesn't react with C2O4, while Ca does in fact react with C2O4
Its formula is NH3, it is basic, and will turn litmus blue
Has water molecule(s) attached ( •H2O).
Precipitates out of solution.
Color litmus paper turns in base.
Color litmus paper turns in acid.
A reagent bottle on the shelf labeled 0.5 NaCl was used in place of the 0.5 M CaCl2. Assuming C2O42- to be in excess, what would be observed as a result of using this wrong reagent in this test?
The Ca2+ would be replaced by Na+ and no precipitate would form = Na2C2O4
Objective Exp. 28: Chemistry of Copper
To observe the chemical properties of copper through a series of chemical reactions
• To determine percent recovery of copper through a cycle of reactions
an element of low percent composition in a mixture of metals, the result of which produces an alloy with unique, desirable properties
substances that readily remove electrons from other substances—i.e., Cu lCu2+ + 2e-
a substance that donates electrons to another substance—i.e., Cu2+ + 2e-lCu
positively charged ions
negatively charged ions
chemical reaction where one reactant is exchanged for one ion of a second reactant
ex. A + BC--> B+ AC
type of reaction where two reactants exchange ions to form two new compounds; typically result in the formation that's a precipitate
ex. AB + CD--> AD+CB
is used to separate the precipitate (solid) from the supernatant (solution)
What is the formula and the color of the gas that is evolved?
the gas was brown
formula: NO2 (Nitrogen dioxide)
When the NaOH solution is added, Cu(OH)2 does not precipitate immediately. What else present in the reaction mixture from Part A reacts with the NaOH before the copper(II) ion?
2O is still present in the mixture. Cu(NO3)2 contains Cu(2+) and NO3(-) ions. When NaOH is added, Na+ ions react with the NO3- ions= NaNO3 and the Cu(2+) ions react with the OH- ions= Cu(OH)2. It's possible that these H+ and OH- ions that form water reacted with NaOH before Cu(NO3)2 did.
The sample in Part B was not centrifuged. Why? Perhaps the student chemist had to be across campus for another appointment. Because of the student's "other priorities" the percent recovery of copper in the experiment will decrease.
Because the sample was not centrifuged, the reaction was not fully finished which means that some copper will remain in the Copper (II) Hydroxide and not react with the NaOH, which ultimately decreases the percent recovery of copper from the experiment.
All of the CuO does not react with the sulfuric acid. Will the reported percent recovery of copper in the experiment be too high or too low?
Because all of the CuO did not react with the sulfuric acid the reported percent recovery of copper will be too low. This is due to the fact that you will only get the part of the CuO that reacted and when doing your calculations you will have a lesser mass of Cu recovered than was expected, making the percent recovery lower than expected.
Sulfuric acid has a dual role in the chemistry of this experiment. What are its two roles in the recovery of the copper metal?
to speed up the reaction in the copper cycle as well as act as a dehydrating agent.
Jacob couldn't find the 6 M H2SO4, so instead substituted the 6 M HNO3 that was available. What change was most likely observed as a result of this decision?
HNO3 oxidizes itself and will not take part in the reaction, meaning that he will not observe any reaction.
Errors in experimental technique can lead to the percent recovery of copper being too high—one such error may occur in Part E.2 and another in Part E.3. Cite those two errors and explain what should be done to ensure that those errors do not occur in the recovery.
failure to centrifuge. To ensure that this error does not occur in recovery you must not let your solution sit idle. You must balance the centrifuge and avoid any contaminants. An error that can occur in part E. 3 is failure to dry the copper completely. Follow technique 13 and heat test tube properly.
What are all those black spots on the ceiling of my lab room?
Probably someone overheated the _____ in their test tube.
What mass of Cu metal is required to react with 4.6 drops of 16M HNO3?
Assume 20 drops = 1 mL. Cu (MM 63.54 g/mol), (16M = 16 mol/L)
Cu(s) + 4HNO3(aq) -->Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
a substance having two protons (H+) that can ionize in an aqueous system
a chemical reaction between an acid and a base
any ionic compound that is a neutralization product of an acid-base reaction.
it exists as ions in solution
it exists as a solid precipitate
solutions have a sour or tart taste, cause a pricking sensation on the skin, and turn blue litmus red; substances that produce hydronium ion, H3O+, in aqueous solutions.
solutions have a bitter taste, are slippery to the touch, and turn red litmus blue; substances that produce hydroxide ion, OH-, in aqueous solutions.
pH less than 7; sulfuric acid, hydrochloric acid, nitric acid, phosphoric acid, acetic acid- vinegar, citric acid- citrus fruits, ascorbic acid- vitamin c
Complete the following molecular equation and identify the Bronsted acid:
NaOH(aq) + HCl(aq) -->
H2O(l) + NaCl(aq); HCl is the acid
Which spectator ions are floating around in solution after the following reaction is complete?
NaOH(aq) + HCl(aq)-->
Hydrochloric acid = strong acid
H+(aq) + Cl-(aq)
Acetic acid = weak acid
CH3COO-(aq) + H+(aq)
Sodium hydroxide = strong base
Na+(aq) + OH-(aq)
ammonia = weak base
NH3(aq) + H2O(aq) <==>
NH4+(aq) + OH-(aq)
cobalt charge in CoCl2
sodium charge in Na2CO3
+1, always +1...this one never changes
copper charge in CuSO4
nickel charge in NiCl2
name for CoCO3
Brønsted Acid-Base Theory
According to this theory, an acid is a "proton donor" and a base is a "proton acceptor."
Name and write the formula of the ion that makes a solution basic?
OH- ; hydroxide ion
Name and write the formula of the ion that makes a solution acidic?
H3O+ ;hydronium ion
Ex. 8 limiting reactant:
How did we filter the product?
Account for % recovery being more or less than 100%. Why?
If you got more than 100% it could be due to extra water or the magnesium strips not completely reacting
a chemical substance is broken down into simpler substances
AB ---> A + B
Two or more chemical species combine to form a more complex product
A + B ---> AB