Ch7 ef review answers
Using this key, pick which category best describes the following chemicals: binary acid, oxy acid, covalent compound ( & not an acid), binary ionic compound (& not acid), or has a polyatomic ion( & not an acid). Then supply either the missing formula or name, & 2nd name for acids & covalent
oxy acid Carbonic acid or hyrdogen carbonate H2CO3
ionic… Magnesium sulfide MgS
polyatomic… Sodium carbonate Na2 CO3
binary acid.. Hl hydrogen iodide or hydroiodidic acid
ionic… CrO Chromium (II) oxide
polyatomic ion…. KOH potassium hydroxide
covalent…. Carbon IV oxide CO 2
oxy acid…. H2SO3 sulfurous acid hydrogen sulfite
binary acid…. Hydrochloric acid HCl hydrogen chloride
covalent….. NO Nitrogen monoxide or Nitrogen (II) oxide
-------------------------------------------------------------------------------------------------------------------------
Give the formulas; for acids and covalent compounds, give a 2nd name
potassium nitride K3 N
nickel III nitride NiN
nitrite NO2-1
hydrogen chlorate HClO3 chloric acid
phosphorus trioxide PO3 phosphorus (VI) oxide
nitrous acid hydrogen nitrite HNO 2
nitrogen VI oxide NO3 nitrogen trioxide
------------------------------------------------------------------------------------------------------------------------ find % sugar in gum from data. 3.00g before, 1.20g after chewing. 3-1.2= 1.8g of sugar;
1.8/ 3.00 = 60%
___________________________________________________
salt water lab also on test! ch7 saltwater d&%
mass of dish 38.0g
mass of dish and 15mL of saltwater NaCl(aq) 59.0g
mass of dish and salt after oven 42.5g
mass of salt 4.5g
mass of water 16.5g
mas of saltwater 21g
density of salt water d=m/v= 21g/15mL = 1.4g/mL
% salt in salt water = salt/saltwater = 4.5/21= 21.4%
_____________________________________________________
A hydrate, barium sulfide triihydrate, is heated, and the following data is obtained.
BaS*3H2O
169 54 total 223
Mass of empty dish 34.06g
Mass of dish and hydrate 37.08g.
Mass of dish and chemical after heating 36.85g.
Mass in g of the hydrate 3.02g
Mass in g of the anhydrous salt 2.79g
Mass in g of the water .23g
% water in the hydrate from the experiment WORK: water/hydrate = .23/ 3.02 = 7.6%
% water in the hydrate from the formula WORK: water/hydrate = 54/223 = 24%
% error WORK: e-a/a = 7.6-24/24 = -68%
moles of water in the hydrate (from data) .23g/18 = .013 mole
moles of anhydrous salt 2.79g / 169 = .0165mole
no, this is about 1:1, we were expecting 3:1
------------------------------------------------------------------------------------------_________________________________________
------------------------------------------------------------------------------------------------------------------------
formula mass of CH4 = 12+1+1+1+1 =16 amu
Common name and important uses (in our lab and in your home) of CH4 methane ; used for bunsen burners; heating the home and water.
Formula mass of Sodium sulfate in amu Na2S04 = 142
% S in Sodium sulfate WORK: 32/142 = 22.5%
% O in Sodium sulfateWORK: 64/142 = 45.1%
% Na in Sodium sulfate WORK: 46/142 = 32.4%
How many moles are 88g of carbon dioxide? WORK: 88/44 =2
What is the mass in grams of 0.55 moles of carbon monoxide? WORK: .55 mole * 28 = 15.4g
mass of 1 mole of sulfur VI oxide in g . SO3= 80g
moles of water present in 162g of water WORK: 162/18 = 9mole
mass of 0.61 mole of water in g WORK: .61mole *18 = 10.98g
------------------------------------------------------------------------------------------
glucose and C2H6 are not empirical; glucose e.f. is CH2O, and C2H6 ef would be CH3
A sample of a compound decomposes into 92.3gC and 15.4gH ; it has a molar mass of 42g.
How many moles of H is 15.4g? WORK: 15.4/1=15.4 moles H
How many moles of carbon is 92.3g? WORK: 92.3/12= 7.7moles C
What is the mole ratio of H to C? WORK:
15.4mole H / 7.7mole C =2/1
What is the empirical formula?CH2
What is the molecular to empirical formula ratio? WORK: 42/14=3/1
What is the molecular formula? 3(CH2) = C3H6
____________________
for more practice: p 253 (e242) #36-39 & p249 empirical and molecular formula problems; see sample problems in the text
_________
-------------------------------
a chemist does not know the formula of a hydrate of sodium sulfate, so he heats the hydrate, obtaining the following data;
empty dish 32.50g
dish and hydrate 35.63g
dish and anhydrous salt 33.83
based on this information, what is the formula of the hydrate? Show work to justify your answer!
Na2SO4= 33.83-32.5=1.33g /142=.00937mole a.s.
H2O = 35.63-33.83 = 1.80g/18 = 0.10mole water
.10mole water/.00937mole a.s. = 10.7 round to 11
therefore Na2SO4 * 11H2O
mg part below, skip for dec 2021
Why does magnesium gain mass when it burns, but wood loses mass when it burns? Oxygen stays with Mg (MgO) and Oxygen leaves in CO2 gas What is rust? iron (III) oxide Would an object appear to gain or lose mass as it rusts?
gain, oxide stays as solid
Some potassium metal is put into a crucible and heated, forming potassium oxide. The following data is collected.
Empty crucible 12.50g
Crucible and metal 14.45g
Crucible and product 15.25g
Assume the reaction is just like the magnesium reaction we did. Write a balanced chemical equation with states:
4K(s) + O2(g) => 2 K2O(s)
2 Mg(s) + O2(g) = 2 MgO(s)
Mass of oxygen in grams = .8g
Mass of metal in grams = 1.95g
Mass of product in grams =2.75g
exp % oxygen in potassium oxide oxygen/potassium oxide = .8/2.75g = 29%
actual % oxygen in potassium oxide (from formula) oxygen/potassium oxide = 16/94 = 17%
% error = e-a/a = 29-17/17 = 70.5%
Moles of oxygen (from data)WORK: .8/16=.05 mole
Moles of metal (from data)WORK: 1.95/39 = .05 mole