Some Things I Like About Myself ( Expository Essay) Essays

Some Things I Like About Myself ( Expository Essay) Essays Some Things I Like About Myself ( Expository Essay) Essay Some Things I Like About Myself ( Expository Essay) Essay † that the possibility to achieve a goal is what makes life meaningful†, this is why I have worked very hard throughout my entire life to achieve the goals and dreams that I have set up for myself , for example my newest goal is to be able to validate my career, in order to accomplish this I need to improve my English skills , for that reason I have returned to school at my age right now I am focusing on oral communication and writing skills. I hope I will not have trouble understanding native speakers and my writing documents will be done in more professional manner. After I have master this language I will be able to get my masters off accounting and work as a CPA, a certify public account, but in order to achieve this result I will to re-take some classes and get new credits, since law changes , as law changes, taxes and other issues are handle different between countries. For many of the other classes I will be able to transfer the credits form my degree in Colombia. Since number does not change, the mathematical problems stay the same. The second quality I have I am friendly person, sometimes when the people come into the my life I know right away that they were meant to be there, to serve some sort of purpose, teach me a lesson, or to help me figure out who I am or who I want to become. You never know who these people may be become to be possibly my neighbor, worker , good friend , lover, or even a complete stranger, when I met some people I sow beyond their physical appearance (poor or rich , black or white , educate or not ) , but I am sure all human being bring to my life positive aspects to learn even though some people are unfriendly and include impolite , but I try to be kind and laid-back to them. A third quality I am prudent for example, I have learned through my many errors in life no to duplicate my unfortunate actions. All my life I have been taught before I act, in order to prevent myself from making mistakes I need to consider the consequences of my actions prior to acting. I guess sometimes I am not spontaneous but I avoid possible future problems for bad actions or decisions. Finally, I recognize I am not perfect, but these qualities as persistent, friendly and prudent brought to my life happiness effect the way I feel about myself and my life and constantly seek to improve myself to become better.

Calculate Empirical and Molecular Formulas

Calculate Empirical and Molecular Formulas The empirical formula of a chemical compound is a representation of the simplest whole number ratio between the elements comprising the compound. The molecular formula is the representation of the actual whole number ratio between the elements of the compound. This step by step tutorial shows how to calculate the empirical and molecular formulas for a compound. Empirical and Molecular Problem A molecule with a molecular weight of 180.18 g/mol is analyzed and found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. How To Find the Solution Finding the empirical and molecular formula is basically the reverse process used to calculate mass percent or mass percentage. Step 1: Find the number of moles of each element in a sample of the molecule.Our molecule contains 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. This means a 100-gram sample contains: 40.00 grams of carbon (40.00% of 100 grams)6.72 grams of hydrogen (6.72% of 100 grams)53.28 grams of oxygen (53.28% of 100 grams) Note: 100 grams is used for a sample size just to make the math easier. Any sample size could be used, the ratios between the elements will remain the same. Using these numbers, we can find the number of moles of each element in the 100-gram sample. Divide the number of grams of each element in the sample by the atomic weight of the element to find the number of moles. moles C 40.00 g x 1 mol C/12.01 g/mol C 3.33 moles C moles H 6.72 g x 1 mol H/1.01 g/mol H 6.65 moles H moles O 53.28 g x 1 mol O/16.00 g/mol O 3.33 moles O Step 2: Find the ratios between the number of moles of each element. Select the element with the largest number of moles in the sample. In this case, the 6.65 moles of hydrogen is the largest. Divide the number of moles of each element by the largest number. Simplest mole ratio between C and H: 3.33 mol C/6.65 mol H 1 mol C/2 mol HThe ratio is 1 mole C for every 2 moles H The simplest ratio between O and H: 3.33 moles O/6.65 moles H 1 mol O/2 mol HThe ratio between O and H is 1 mole O for every 2 moles of H Step 3: Find the empirical formula. We have all the information we need to write the empirical formula. For every 2 moles of hydrogen, there is one mole of carbon and one mole of oxygen. The empirical formula is CH2O. Step 4: Find the molecular weight of the empirical formula. We can use the empirical formula to find the molecular formula using the molecular weight of the compound and the molecular weight of the empirical formula. The empirical formula is CH2O. The molecular weight is molecular weight of CH2O (1 x 12.01 g/mol) (2 x 1.01 g/mol) (1 x 16.00 g/mol)molecular weight of CH2O (12.01 2.02 16.00) g/molmolecular weight of CH2O 30.03 g/mol Step 5: Find the number of empirical formula units in the molecular formula. The molecular formula is a multiple of the empirical formula. We were given the molecular weight of the molecule, 180.18 g/mol. Divide this number by the molecular weight of the empirical formula to find the number of empirical formula units that make up the compound. Number of empirical formula units in compound 180.18 g/mol/30.03 g/molNumber of empirical formula units in compound 6 Step 6: Find the molecular formula. It takes six empirical formula units to make the compound, so multiply each number in the empirical formula by 6. molecular formula 6 x CH2Omolecular formula C(1 x 6)H(2 x 6)O(1 x 6)molecular formula C6H12O6 Solution: The empirical formula of the molecule is CH2O.The molecular formula of the compound is C6H12O6. Limitations of the Molecular and Empirical Formulas Both types of chemical formulas yield useful information. The empirical formula tells us the ratio between atoms of the elements, which can indicate the type of molecule (a carbohydrate, in the example).  The molecular formula lists the numbers of each type of element and can be used in writing and balancing chemical equations. However, neither formula indicates the arrangement of atoms in a molecule. For example, the molecule in this example, C6H12O6, could be glucose, fructose, galactose, or another simple sugar. More information than the formulas is needed to identify the name and structure of the molecule. Empirical and Molecular Formula Key Takeaways The empirical formula gives the smallest whole number ratio between elements in a compound.The molecular formula gives the actual whole number ratio between elements in a compound.For some molecules, the empirical and molecular formulas are the same. Usually, the molecular formula is a multiple of the empirical formula.