Chemistry (Math in Disguise of Science) – Flashcards
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Calculate the mass in grams 1.25 moles of aluminum chloride.
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1. Add the atomic mass, multiply them if they have a little number. 2. Multiply the number they gave you over 1 GRAM and the number you got from adding the masses over 1 MOLE. EX Al-26.98 Cl-(3)35.45 133.33 1.25 mol/1 g X 133.33 g/1 mol= 166.62 g /mol
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Calculate the number of molecules present in 4.75 g of phosphine , PH3(little 3).
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1. Add the atomic mass, multiply them if they have a little number. 2. Multiply the number they gave you over 1 MOLECULE and the AVAGOTES number over the number you got. EX P-30.97 H-(3)1.0079 33.99 g 4.75g/1 Molecules X 6.22x10^23/33.99 g = 8.42x10^23 Molecules
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Calculate the percent by mass of Methane CH4(little 4).
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1. Add the atomic mass, multiply them if they have a little number. 2. Atomic mass g over the number you got. 3. Times 100 EX C-12.01 H-(4)1.0079 16.04 g C- 12.01gC/16.04g X 100= 75% H- 4.03gH/16.04g X 100=25%
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A compound was analyzed and found to contain the following . Baruim-89.59% Oxygen-10.44% Determine the empirical formula of it.
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1. Change the percent to grams. 2. Put the number they gave you over 1 MOLE and multiply it by 1 MOLE over its atomic mass. 3. Put the number you got over the smallest number. 4. The number you now have is the subscript. EX Ba-89.59%=89.59g O-10.44%=10.44g 89.59g Ba/ 1 mol X 1 mol/ 137.3 g = 0.6253 mol Ba 10.44g O/ 1 mol X 1 mol/16.00 g = 0.6253 mol O 0.6253 mol Ba/0.6253 mol =1 mol Ba 0.6253 mol O/0.6253 mol = 1 mol O = BaO
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An oxide of aluminum if formed by the reaction of 4.15 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula for this compound.
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1. Put the number they gave you over 1 mole times 1 mol over the atomic mass. 2. Divide both by the smallest number. 3. If you don't get a full number , than multiply ALL till you do. EX Al- 4.151g Al/1 mol X 1 mol/26.98 g Al = .1538 O- 3.692g O/1 mol X 1 mol/ 16 G O= .23075 .1538/ .1538 = 1 X 2= 2 .23075/.1538=1.5 X 2=3 =Al2O3 ( Little 2 & Little 3)
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A compound with the empirical formula CH4O (little 4) was formed in a subsequent experiment to have a molar mass of apprimatldy 192 g. What is the molecule formula of the compound?
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1. Add the atomic mass, multiply them if they have a little number. 2. Put the number they gave you over the number you got. 3. If you got number that info full multiply it till you do. EX C-12.01 H-(4)1.008 O-15.9994 32.0414 192g/32.0414g = 5.99= 6 C6H24O6 (little 4, little 24, little 6)
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Calculate the number of moles in 6.22 g of Aluminum.
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1. Put the number they gave you over 1 mole and times it by 1 mole over the atomic mass EX 6.22 g Al/ 1 mol X 1 mol/ 26.98 g Al = 0.23 g/mol Al
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Calculate the molar mass for magnesium nitride, Mg(NO2)2 ( little 2 ,little 2).
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1. Add the atomic mass, multiply them if they have a little number. EX Mg-24.305 N-(2)14.00674 O-(4)15.9994 116.31
