Hello i need assistance. i have some formulas but am uncertain as to how to proc

Question

Hello i need assistance. i have some formulas but am uncertain as to how to proceed. like the first question. it has a formula but where does the +6 come from and would i need this to work a-c?

1. Calculate (to the nearest 0.1 u) the formula mass of these compounds.
(a) aluminum peroxide, Al2(O2)3 ________u 2*(mass of Al)+6*(mass of O)...do this for all of these
(b) rubidium nitrate, RbNO3___________u
(c) carbon dioxide, CO2__________u


2. Find the percentage by mass of Cl in MgCl2 if it is 25.1% Mg by mass. _____%
If it is 25.1 % by mass Mg, then it is 100-25.1% Cl by mass because there is only Mg and Cl in this compound
3.




Determine the percentage by mass of each element in the following compounds.
(a) salt, NaCl do this:
Na_________% (mass of Na)/(mass of NaCl)
Cl_________% (mass of Cl)/(mass of NaCl)

(b) grain alcohol, C2H5OH
C_______% do the same thing as above
H_______%
O_______%




4. In a lab experiment, 7.35 g of phosphorus reacts with bromine to form 64.19 g of phosphorus tribromide.
(a) Calculate the percentage by mass of P in phosphorus tribromide._______% (mass of P)/(mass of phosphorus tribromide) mass of phosporous tribromide=mass of phosporous+3*mass of bromine
(b) How many grams of bromine reacted?_______g There are only two elements present: phosphorus and bromine. In the end we have 64.19 g of PBr3 and 7.35 g of that is P, so the rest is bromine. So the grams of bromine is 64.19g -7.35g
How do you know? ________
law of constant composition, law of conservation of mass, law of multiple proportions, or law of definite proportions?


5. In a lab experiment, 7.75 g of phosphorus reacts with bromine to form 67.68 g of phosphorus tribromide. How much phosphorus tribromide would be formed if 10.00 g of phosphorus reacted with 61.10 g of bromine?_______g
How do you know?
law of constant composition Dalton's law law of definite proportions law of conservation of mass ?

Explanation / Answer

You should post them separately. For question 1: a) Al2(O2)3 ... Determine how many aluminum and oxygen you have. The two behind aluminum = 2 Al There are two Oxygen, but you have to multiply the 3 that is on the outside to this. 2*3 = 6 oxygen Al = 26.98 u * 2 Al = 53.96 u O = 15.9994 u * 6 = 95.9964 u Add them up: 53.96+95.9964 = 149.956 ... Round: 150.0 u b) RbNO3 You have 1 RB and 1 N and 3 O (the three behind O indicates this). Rb = 85.4678 u N = 14.0067 u O = 15.9994 * 3 = 47.9982 Add them up: 85.4678+14.0067+47.9982= 147.473 ... Round: 147.5 u c) CO2 You have 1 C and 2 O (noticed the 2 behind O). C = 12.0107 O = 15.9994 * 2 = 31.9988 u Add up : 12.0107+31.9988= 44.0095 ..... round: 44.0 u

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