Carbon dioxide formula mass

Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is 1.68 g. From the data given, show that the simplest formula of the oxide is X2O3 and write a balanced equation for the reaction. 

Dimefoylhydrazine, foe fuel used in foe Apollo lunar descent module, has a molar mass of 60.10 g/mol. It is made up of carbon, hydrogen, and nitrogen atoms. The combustion of 2.859 g of the fuel in excess oxygen yields 4.190 g of carbon dioxide and 3.428 g of water. What are foe simplest and molecular formulas for dimefoylhydrazine ... 

A combustion analysis was carried out on 1.621 g of a newly synthesized compound, which was known to contain only C, H, and O. The masses of water and carbon dioxide produced were 1.902 g and 3.095 g, respectively. What is the empirical formula of the compound ... 

In a combustion analysis, the amounts of C, H, and O atoms in a sample of a compound, and thus its empirical formula, are determined from the masses of carbon dioxide and water produced when the compound bums in excess oxygen. 

M.9 The stimulant in coffee and tea is caffeine, a substance of molar mass 194 g-mol When 0.376 g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.1 10 g of nitrogen were formed. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion. 

M.10 Nicotine, the stimulant in tobacco, causes a very complex set of physiological effects in the body. It is known to have a molar mass of 162 g-mol. When a sample of mass 0.385 g was burned, 1.072 g of carbon dioxide, 0.307 g of water, and 0.068 g of nitrogen were produced. What are the empirical and molecular formulas of nicotine Write the equation for its combustion. 

Compounds that do not decompose cleanly into their elements must be analyzed by other means. Combustion analysis is particularly useful for determining the empirical formulas of carbon-containing compounds. In combustion analysis, an accurately known mass of a compound is burned in a stream of oxygen gas. The conditions are carefully controlled so that all of the carbon in the sample is converted to carbon dioxide, and all of the hydrogen is converted to water. Certain other elements present in the sample are also converted to their oxides. 

C03-0089. Combustion analysis of 0.60 g of an unknown organic compound that contained only C, H, and O gave 1.466 g of carbon dioxide and 0.60 g of water in a combustion analysis. Mass spectral analysis showed that the compound had a molar mass around 220 g/mol. Determine the empirical formula and molecular formula. 

M.9 In a combustion analysis of a 0.152-g sample of the artificial sweetener aspartame, it was found that 0.318 g of carbon dioxide, 0.084 g of water, and 0.0145 g of nitrogen were produced. What is the empirical formula of aspartame The molar mass of aspartame is 294 g-mol. What is its molecular formula ... 

The chemical symbol for an element—H, C, O, etc.—is used to designate that element. Molecular substances consist of independent molecules containing two or more atoms bound together. A molecular formula specifies the identity and number of the atoms in the molecule. For instance, the formula for carbon dioxide is C02, one carbon atom and two oxygen atoms. The molecular mass of C02 is Ar(C) + 2Ar(0) = 12.0107 + 2 x... 

Each element has a specific symbol that is different from the symbol for any other element. In a chemical formula, the symbol stands for an atom of an element. Molecular substances are composed of two or more atoms that are tightly bound together. The formula for a molecular substance consists of the symbols for the atoms that are found in that molecule. For instance, the formula for carbon dioxide is CO2. Note the use of the subscript to show that each molecule contains two oxygen atoms in addition to the one carbon atom. Also note that the 1 for the one carbon atom is not written. The molecular mass of CO2 is the sum of the atomic mass of carbon plus twice the atomic mass of oxygen and is expressed in u. As was discussed directly above, the molar mass of CO2 is the mass in grams equal to the molecular mass in u. A mole of carbon dioxide is 12.0 u + 2(16.0 u) = 44 u. This result can be expressed as 44 g to indicate one Avogadro s number, Na, of CO2 molecules. Recall that Na is 6.0221 x 1023 things—molecules in this case. 

Chemical formulas such as CO and C02 reflect an important law called the law of multiple proportions. This law applies when two elements (such as carbon and oxygen) combine to form two or more different compounds. In these cases, the masses of the element (such as 02 in CO and C02) that combine with a fixed amount of the second element are in ratios of small whole numbers. For example, two moles of carbon can combine with one mole of oxygen to form carbon monoxide, or with two moles of oxygen to form carbon dioxide. The ratio of the two different amounts of oxygen that combine with the fixed amount of carbon is 1 2. 

Similarly, all the carbon in the sample has been incorporated into the carbon dioxide. Multiply the mass percent (as a decimal) of carbon in carbon dioxide by the mass of the carbon dioxide to get the mass of carbon in the sample. Convert to moles and determine the empirical formula. 

An 874 mg sample of cortisol was subjected to carbon-hydrogen combustion analysis. 2.23 g of carbon dioxide and 0.652 g of water were produced. The molar mass of cortisol was found to be 362 g/mol using a mass spectrometer. If cortisol contains carbon, hydrogen, and oxygen, determine its molecular formula. 

In section 6.4, you learned several practical methods for determining empirical and molecular formulas of compounds. You may have noticed that these methods work because compounds react in predictable ways. For example, you learned that a compound containing carbon and hydrogen reacts with oxygen to produce water and carbon dioxide. From the mass of the products, you can determine the amount of carbon and hydrogen in the reactant. You also learned that a hydrate decomposes when it is heated to form water and an anhydrous compound. Again, the mass of one of the products of this reaction helps you identify the reactant. In Chapter 7, you will learn more about how to use the information from chemical reactions in order to do quantitative calculations. 

When platinum is heated to whiteness with yellow phosphorus in a Hessian crucible, fusion takes place and, when the excess of phosphorus has burned off, a brittle mass of empirical formula PtgP5 is obtained on cooling.1 The same substance lesults 2 on heating platinum in phosphorus vapour in a current of carbon dioxide. The temperature is raised to the point necessary to start the reaction, and the friable residue of Pt3P5 remains behind. 

Emergence of chemical atomism. The beginning of the 19th century was marked by the emergence of atomic theory developed by the famous English scientist John Dalton (1766-1844). According to Dalton, atoms combine in multiple proportions. Thus, if two atoms form only one compound, they associate in the 1 1 ratio if two atoms form two compounds, their ratios in these compounds are respectively 1 1 and 1 2, and so on. Accordingly, Dalton adopted the formula HO for water, CO for carbon monoxide, COg for carbon dioxide, and so on. From these formulas he determined the atomic masses of elements. It is clear that frequently both atomic masses and chemical formulas of Dalton were not correct. 

Now, let s see an example involving the formula mass of carbon dioxide. 

The empirical formula of a hydrocarbon can be determined using an instrument similar to the one shown in Figure 1. A sample hydrocarbon is combusted. The absorption chambers absorb all the water and carbon dioxide from the reaction. CaCl, can absorb both water and CO,. The masses of the chambers before and after the reaction are compared to find the moles of carbon and hydrogen in the sample. 

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