# Round all masses to the 10ths place  Step One: two atoms of iron and three of oxygen  Step Two: Fe = 55.8 amu and O = 16.0 amu  Step Three: (2 x 55.8)

## Presentation on theme: "Round all masses to the 10ths place  Step One: two atoms of iron and three of oxygen  Step Two: Fe = 55.8 amu and O = 16.0 amu  Step Three: (2 x 55.8)"— Presentation transcript:

Round all masses to the 10ths place

 Step One: two atoms of iron and three of oxygen  Step Two: Fe = 55.8 amu and O = 16.0 amu  Step Three: (2 x 55.8) + (3 x 16.0)  Step Four: 111.6 + 48.0 = 159.6 amu (grams) for one mole

 Step One: one atom of potassium, one of chlorine, and four of oxygen.  Step Two: K = 39.1 amu, Cl = 35.5 amu, O = 16.0 amu  Step Three: (1 x 39.1) + (1 x 35.5) + (4 x 16.0)  Step Four: 39.1 + 35.5 + 64.0 = 138.6 amu (grams) for one mole

 Step One: two atoms of nitrogen, eight of hydrogen and one of sulfur  Step Two: N = 14.0 amu, H = 1.0 amu, S = 32.1 amu  Step Three: (2 x 14.0) + (8 x 1.0) + (1 x 32.1)  Step Four: 28 + 8.0 + 32.1 = 68.1 amu (grams) for one mole

 Calculate the molar mass of Al(NO 3 ) 3  Al(1 x 27.0) + N(3 x 14.0) + O(9 x 16.0) =  213.0 amu (grams) is the mass of one mole of aluminum nitrate.

 1. AlCl 3  2. Ba(SCN) 2  3. LiH  4. Ba(BrO 3 ) 2  5. AlBr 3  6. HCl

 1. 133.5 g/mol  2. 253.5 g/mol  3. 7.9 g/mol  4. 393.1 g/mol  5. 266.7 g/mol  6. 36.5 g/mol

Round all final answers to 3 sig figs

 KMnO 4 = 158.0 g / mole of KMnO 4  Now let’s solve it  25.0 g x 1 mole = 0.158191804 moles 158.0 g  0.158 moles

 H 2 O 2 = 34.0 g / mole  Now, let’s solve it  17.0 g x 1 mol H 2 O 2 = 0.5 moles 34.0 g  0.500 moles

 Calculate the moles present in:  1) 2.00 grams of H 2 O  2) 75.57 grams of KBr  3) 100. grams of KClO 4  4) 8.76 grams of NaOH  5) 0.750 grams of Na 2 CO 3  Remember to round answers to 3 sig figs

 1)0.111 moles  2)0.635 moles  3) 0.722 moles  4)0.219 moles  5)0.00708 moles

Round all answers to 3 sig figs

 H 2 O 2 molar mass = 34.0 grams/mole  0.700 moles x 34.0 grams = 23.8 g 1 mole  23.8 grams

 KClO 3 molecular mass = 122.6 g/mole  2.50 moles x 122.6 g = 306.5 grams 1 mole  307 grams

 Calculate the grams present in:  1) 0.200 moles of H 2 S  2) 0.100 moles of KI  3) 1.500 moles of KClO  4) 0.750 moles of NaOH  5) 3.40 x 10 -5 moles of Na 2 CO 3  Remember to round answer to 3 sig figs

 1)6.82 grams  2)16.6 grams  3)136 grams  4)30.0 grams  5)0.00360 grams

6.02 x 10 23 particles = 1 mole Particles can mean atoms or molecules

 One mole of donuts contains 6.02 x 10 23 donuts  One mole of H 2 O contains 6.02 x 10 23 molecules  One mole of nails contains 6.02 x 10 23 nails  One mole of Fe contains 6.02 x 10 23 atoms

 One mole of dogs contains 6.02 x 10 23 dogs  One mole of electrons contains 6.02 x 10 23 electrons  One mole of Chemistry students contains 6.02 x 10 23 poor, suffering (I mean happy, joyful) high school students

 0.450 mole Fe x 6.02 x 10 23 atoms 1 mole  = 2.709x10 23 atoms  Final, rounded answer: 2.71x10 23 atoms

 0.200 mole x 6.02 x 10 23 atoms 1 mole  = 1.204x10 23 atoms  Final, rounded answer: 1.20x10 23 atoms

 1) Calculate the number of molecules in 1.058 mole of H 2 O  2) Calculate the number of atoms in 0.750 mole of Fe  Remember to round to 3 sig figs

 1)6.37x10 23 molecules  2)4.52x10 23 atoms

Grams to particles OR particles to grams Grams to atoms OR atoms to grams If MOLE is not stated, it’s a two step problem ROUND ALL ANSWERS TO 3 SIG FIGS

 0.450 g x 1 mole Fe x 6.02x10 23 atoms = 55.8 g 1 mole  4.8548387x10 21 atoms  Final answer = 4.85x10 21 atoms

 H 2 O molecular mass = 18.0 g  0.200 g x 1 mole H 2 O x 6.02x10 23 molecules 18.0 g 1 mole  6.6888888889x10 21 molecules  Final answer = 6.69x10 21 molecules

 1) Calculate the number of molecules in 1.058 gram of H 2 O  2) Calculate the number of atoms in 0.750 gram of Fe  Remember to round answer to 3 sig figs

 1)3.54x10 22 molecules  2)8.10x10 21 atoms

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