Equivalent Mass = Mass of the Metal × 8 Mass of Oxygen in the Oxide. Fertilizer grades of 45-15-30, 36-12-24, and 9-3-6 also have a 3-1-2 ratio. The ratio of the fertilizer in the example above is 3-1-2. That is, while element masses differ, when it comes to bonding with other atoms, the number of atoms, expressed in moles, is the determining factor in how much of a given element or compound will react with a given mass of another. 5) (CAS No. The concept of equivalent weight allows you to explore the fact that atoms combine to form molecules in fixed number ratios, not mass ratios. Compare this with the manufacturer's specification. = 122.5) will be (K C I O 3 → K C I + O 2 ) View solution Number of g of oxygen in 3 … From your mass of MgNH4PO4.6H2O collected, calculate your value a) for the mass of "P2O5" in the initial mass of fertilizer analyzed b) the % "P2O5" in the fertilizer. Equivalent weight (also known as gram equivalent) is the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance.The equivalent weight of an element is the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. A given weight of two fertilizers can contain different amounts of actual plant food. Fertilizer grades of 45-15-30, 36-12-24, and 9 … The ratio of the fertilizer in the example above is 3-1-2. Again, using 10-20-10 as an example, the ratio is 10/10-20/10-10/10 = 1-2-1. A fertilizer with a 3- 1-2 ratio contains one and a half times as much N as K2O and three times more N than P2O5. The percent loss in weight after heating a pure sample of potassium chlorate (mol. O. VANADIUM PENTOXIDE (V. 2. A fertilizer with a 3-1-2 ratio contains one and a half times as much N as K2O and three times more N than P2O5. One gram of hydrogen is found to combine with 8 0 g of bromine one gram of calcium valency = 2 combines with 4 g of bromine the equivalent weight of calcium is View solution Calculate the weight of C a O that can be obtained by heating 2 0 0 k g of limestone which is 9 5 % pure. Body weight gain and lung weights in rats (F344/N) exposed to vanadium pentoxide by inhalation for 3 months (values are means ± standard . Suppose you analyze a sample of fertilizer by the procedure of this experiment and obtain 1.000 g of MgNH4PO4 x 6H20(s) as the product. O by weight. the relationship between the N - P2O5 - and K2O content of a fertilizer. \text{Equivalent Mass}=\dfrac{\text{Mass of the Metal}\times 8}{\text{Mass of Oxygen in the Oxide}}. Could someone please help me with this? I don't understand where the P205 came from. $\ce{P2O5}$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. I'm thinking the oxygen is a spectator ion or something? wt. Calculate the equivalent mass of P2O5 initially in the sample. This might seem silly given that as you pointed out anything from $0-70\% \ \ce{P2O5}$ could just be expressed as a percentage of $\ce{H3PO4}$ but there are times when a higher $\ce{P2O5}$ concentration is desired. The mass of a metal in its oxide is 75 % 75\% 7 5 % of the mass of the whole compound. Fertilizer ratio is the ratio of weight percents of N-P 2 O 5-K 2 O and is determined by dividing the three numbers by the smallest of the three. The ratio of a fertilizer is the relationship between the N - P2O5 - and K2O content of a fertilizer. Equivalent Mass = Mass of Oxygen in the Oxide Mass of the Metal × 8 . Just want to know how i would go about this question, where would I start. ... Human equivalent dose conversion by BW3/4 for RfD derivation.....76 Table 5-3. 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