Chem 130: Balancing Equations, Basic Calculations

A bracelet contains 14.1 g of silver and an unknown amount of copper. The total mass of the bracelet is 17.6 g. Calculate the % silver by mass and find the % of copper in the bracelet.

% Silver by mass: 14.1g/17.6g x 100% = 80.1% silver Because the substance is 80.1% silver, the remaining amount is copper, hence 100- 80.1% = 19.9% copper

Balance the equation to produce dimethylhydrazine and dinitrogen tetraoxide used in the moon lander (1969) and space shuttles (CH3)2N2H2 (l) + N2O4 (l) ---> N2 (g) + H2O (l) + CO2 (g)

(CH3)2N2H2 (l) +2 N2O4 (l) --->3 N2 (g) + 4 H2O (l) + 2 CO2 (g)

Define density

mass in g/volume (mL)

How to calculate formal charge

[# of valence electrons on atom] - [non-bonded electrons + number of bonds].

Balance the equation: Al (s) + O2 (g) ---> Al2O3 (s)

4 Al (s) + 3 O2 (g) ----> 2 Al2O3 (s)

Balance and write the net ionic equation: HCl (aq) + MnO2 (s) ----> MnCl2 (aq) + Cl2 (g) + H2O (l)

4 HCl (aq) + MnO2 (s) ----> MnCl2 (aq) + Cl2 (g) + 2 H2O (l) 4 H+ (aq) + 2 Cl- (aq) + MnO2 (s) ----> Mn2+ (aq) + Cl2 (g) + 2 H2O (l)

Convert 37 Celsius (body temperature) to Kelvins

K= C + 273.15 K 37 C + 273.15 K = 310.15 K

Find how many water molecules are in CoSO4 * x H2O (cobalt II sulfate x hydrate). You weighed 1.023 g of the solid and then heated it in a crucible. After the water was driven out, you are left with 0.603 g of blue anhydrous cobalt (III) sulfate. Write the equation.

1) 1.023 g of CoSO4 * x H2O + heat ----> 0.603 g CoSO4 + ?g H2O 2) Because mass is conserved 0.420 g of water must have been driven out (1.023 g - 0.603 g) 3) Now convert masses to moles 0.420 g H2O x 1 mole H2O/18.02 g H2O = 0.0233 moles H2O 0.603 g CoSO4 x 1 mole CoSO4/155.0g CoSO4 = 0.00389 moles of CoSO4 4) Determine the value of x by calculating the mole ratio: moles of H2O/moles of CoSO4= 0.0233 moles H2O/0.00389 moles CoSO4 = 5.99 moles H2O/1.00 mole CoSO4. This tells us that the water to CoSO4 ration is 6:1 so the formula is then CoSO4 *6 H2O. The name is: Cobalt (II) sulfate hexahydrate.

Calculate the mass of one molecule of ammonia NH3.

1) NH3 has a molar mass (MM)= 17.03 g/mole 2) 17.03 g/mole of NH3 x 1 mole/6.022 x 10^23 molecules= 2.828 x 10^-23 grams.

Analysis shows that 0.586 g of potassium can combine with 0.480 g of O2 gas to give a white solid ionic compound with a formula KxOy. What is the formula of the white solid?

1) We find the number of moles of K and O 0.586 g K x 1 mole K/39.10 g K = 0.0150 moles K 0.480 g O2 x 1 mole O2/32.00 g O2 x 2 moles O/1 mole O2= 0.0300 moles O 2) Ratio of moles: 0.0300 moles O/0.0150 moles K = 2/1 3) The empirical formula of the compound is KO2 (potassium superoxide K+ O2-)

Vanillin is a common flavoring agent. Its molar mass is 152 g/mole and is 63.15% C and 5.30% H, the remainder is oxygen. What are the empirical and molecular formulae of vanillin?

1) We find the number of moles of each element in a 100.00 g sample of vanillin (100 g is simpler to use for %) 63.15 g of C x 1 mole C/12.011 g C = 5.258 moles C 5.30 g of H x 1 mole of H/1.008 g of H = 5.26 moles H 2) Mass of oxygen: Since the total mass was set at 100.00 g then we can write: 63.15% of C = 63.15 g C/100.00 g of vanillin 5.30% of H = 5.30 g H/100.00 g of vanillin so 100.00 g of vanillin = 63.15 g C + 5.30 g H + x grams of O so grams of O = 31.55 g 3) Calculate the moles of O 31.55 g O x 1 mole of O/15.999 g of O = 1.972 moles O 4) Find the mole ratios by basing the ratios on the element with the smallest number of moles present, in this case, it is oxygen. mole C/mole O= 5.258 moles C/1.972 moles O= 2.666 moles C/1.000 moles O= 8/3 mole H/mole O = 5.26 moles H/1.972 moles O= 2.67 moles H/1.00 moles O = 8/3 The empirical formula is C8H8O3, MM of 152 g/mole which corresponds to the experimentally determined molar mass so the molecular formula is the same as the empirical formula.

Define mole and give the value of Avogadro's number

A mole of a substance is the its amount that contains as many elementary entities (atoms, molecules, particles etc...) as there are atoms in exactly 12 grams of the carbon-12 isotope. One mole always contains the same number of particles regardless of the substance! 1 mole = 6.022 x 10^23 particles

Balance the reaction for the combustion of butane C4H10.

C4H10 (g) + O2 (g) ----> CO2 (g) + H2O (l) (unbalanced) 1) Balance the C atoms (4C = 4 CO2) 2) Balance the H atoms (10 H on the left and 5 H2O on the right) 3) Balance the O atoms (13 O on the right and 2 on the left) (see below on how to work this out) Solution 1: to have 13 atoms on the left side, use a stoichiometric coefficient of 13/2, you can do this because 13/2 x 2 oxygen atoms/1 molecule O2 = 13 oxygen atoms C4H10 (g) + 13/2 O2 (g) ----> 4 CO2 (g) + 5 H2O (l) Solution 2: Multiply each coefficient after step 3 so that there is an even number of oxygen atoms on the right side 2 C4H10 (g) + 13 O2 (g) ----> 8 CO2 (g) + 10 H2O (l)

Balance and write the net ionic equation: CdCl2 (aq) + NaOH (aq) ----> Cd(OH)2 (s) + NaCl (aq)

CdCl2 (aq) + 2 NaOH (aq) ----> Cd(OH)2 (s) + 2 NaCl (aq) Cd2+ (aq) + 2 OH- (aq) ----> Cd(OH)2 (s)

Balance the equation to produce hydrazine N2H4 a good industrial reducing agent: H2NCl (aq) + NH3 (g) ---> NH4Cl (aq) + N2H4 (aq)

H2NCl (aq) + 2 NH3 (g) ---> NH4Cl (aq) + N2H4 (aq)

Balance the equation: UO2 (s) + HF (l) ----> UF4 (s) + H2O (l)

UO2 (s) + 4 HF (l) ----> UF4 (s) + 2 H2O (l)