balancing chemical reactions

This is an example of an unbalanced chemical reaction. There are two oxygen atoms on the left but only one on the right. In next slide we show procedure to balance chemical reactions. H2OH2O H 2 O 2 +

Procedure for balancing chemical reactions: 1.Choose the molecule with the greatest number of elements. Write in pencil a 1 before this molecule. 2. Find an element in this compound which appears only once more in the reaction. Balance for this element. Write coefficient in pencil. 3. Find another element which appears once more and balance for this element. Reiterate until all molecules have a coefficient. 4. Multiply all coefficients by divisors's least common multiple. 1 H 2 O H2 H2 O H 2 O H2 H2 1/2 O H 2 O 1H2 1H2 1/2 O H 2 O 2H22H2 1 O 2 +

NH 3 + O 2 → N 2 + H 2 O Please balance the following equation

atomic and molecular mass

Concept 1:There are 90 or so naturally occuring atoms. They are the elements.

Concept 2: The different elements/atoms all have more or less a circular shape. A good analogy is eggs. Egg tradition is to count the eggs by the dozen

A dozen hummingbird eggs weigh 12 grams A dozen quail eggs weigh 140 grams. A dozen chicken eggs weigh 600 grams A dozen turkey eggs weigh 1000 grams A dozen ostrich eggs weigh 10,000 grams 1 Hu 12 g/dz 6 Qu 140 g/dz 20 Ch 600 g/dz 30 Tu 1000 g/dz 100 Os g/dz

1 Hu 12 g/dz 6 Qu 140 g/dz 20 Ch 600 g/dz 30 Tu 1000 g/dz How much does a dozen chicken eggs weigh? How much do 3 dozen quail eggs weigh? If I have 2400 grams of chicken eggs how many dozens of chicken eggs do I have? If I have 2500 grams of chicken eggs how many dozens of chicken eggs do I have?

B is for boron. Boron is the fifth smallest element. We count atoms not by the dozen but by the mole. There are grams per mole of boron atoms.

Atoms are very small. A dozen atoms would not weigh any appreciable amount. A million atoms is too small a number to be weighed by any ordinary machine. Nor could a trillion or a quadrillion atoms be weighed an ordinary machine either. The problem with counting atoms We can weigh 100,000,000,000,000,000,000,000 atoms. A mole is 602,214,129,000,000,000,000,000.

couple=2 dozen=12 gross=144 ream=500 mole= 602,214,129,000,000,000,000,000. Words used in English to count groups of people or objects

How much does one mole of hydrogen atoms weigh? How much do two moles of bismuth (Bi) atoms weigh? If I have 10 grams of carbon atoms, how many moles do I have? How much does two mole of H atoms together with one mole of oxygen atoms weigh?

How much does one mole of hydrogen atoms weigh? How much do two moles of bismuth (Bi) atoms weigh? If I have 10 grams of carbon atoms, how many moles do I have? How much does two mole of H atoms together with one mole of oxygen atoms weigh? The molecular weight or MW is the weight of a mole of the molecule.

What is the molecular weight of NH 3 ? How much does two moles of NH 3 weigh? What is the MW of molecular oxygen? How much does 2.5 moles of oxygen molecules weigh? If I have 10 grams of molecular oxygen, how many moles do I have?

combustion analysis

The molecular formula of ethane is C 2 H 6. ethane molecular and empirical formula the empirical formula is the molecular formula coefficients divided by these coefficient’s greatest common divisor. The empirical formula of ethane is CH 3. Two types of chemical formula

We determine the empirical formula by combustion analysis.

an unknown organic compound containing C, H and O (for simplicity ignore N or other elements for now) CnHmOpCnHmOp the empirical formula of an unknown organic compound Goal: determine values of n, m, and p.

The unbalanced chemical reaction for burning is: C n H m O p + O 2 CO 2 + H 2 O We find n, m, and p by burning the compound For this course, burning and combustion mean the same thing.

unbalanced reaction C n H m O p + O 2 CO 2 + H 2 O 1 C n H m O p + O 2 n CO 2 + m/2 H 2 O complicated number of balanced reaction

Strategy for combustion analysis strategy (part 1) 1.Weigh the compound before burning and weigh the carbon dioxide and water after burning. 2. Use number of grams of CO 2 and H 2 O to figure out the number of moles of CO 2 and H 2 O. These values will get us values n and m. 3. Use initial weight of compound and n and m to get p. C n H m O p + O 2 n CO 2 + m/2 H 2 O complicated number of

Strategy for combustion analysis strategy (part 2) 1.Weigh the compound before burning and weigh the carbon dioxide and water after burning. 2. Use number of grams of CO 2 and H 2 O to figure out the number of moles of CO 2 and H 2 O. These values will get us values n and m. 3. Use initial weight of compound and n and m to get p. C n H m O p + O 2 n CO 2 + m/2 H 2 O complicated number of

1. Weigh the original C n H m O p sample. 2. Burn sample compound and weigh in grams the H 2 O and CO 2 produced by burn. 3. Convert grams of H 2 O and CO 2 to number of moles of H 2 O and CO Number of moles of CO 2 is n. Number of moles of H 2 O is m/2. 5. Calculate the weight in grams of m H and and n C. 6.Compare weight from (5) to weight determined in (1). Difference is weight of oxygen atoms. 7.Convert weight of oxygen atoms in (6) to number of moles of oxygen atoms. This number is p. 8. Use n, m. and p and knowledge of fractions to deduce the empirical chemical formula. Procedure for combustion analysis problems CnHmOpCnHmOp C n H m O p + O 2 n CO 2 + m/2 H 2 O complicated number of

A 30.5 g sample of a pure compound containing only C, H, and O is burnt in a combustion analysis apparatus. 66 g of carbon dioxide and 40.5 g of water are accumulated. What is the empirical formula of this compound? 1. Weigh the original C n H m O p sample. 2. Burn sample compound and weigh in grams the H 2 O and CO 2 produced by burn. 3. Convert grams of H 2 O and CO 2 to number of moles of H 2 O and CO Number of moles of CO 2 is n. Number of moles of H 2 O is m/2. 5. Calculate the weight in grams of m H and and n C. 6.Compare weight from (5) to weight determined in (1). Difference is weight of oxygen atoms. 7.Convert weight of oxygen atoms in (6) to number of moles of oxygen atoms. This number is p. 8. Use n, m. and p and knowledge of fractions to deduce the empirical chemical formula.